A 


/«(lH^<'4/^ 


I'-^-L'VL/Vi..^^ 


V. 


\ 


TREATISE 

.  OF 

.ALGEBRA: 

WHEREIN    THE 

PRINCIPLES  ARE  DEMONSTRATED, 

AND     APPLIED 

IN  MANY  USEFUL  AND  INTERESTING  INQUIRIES,  AND  IN 
THE  RESOLUTION  OF  A  GREAT  VARIETY  OF  PROBLEMS 
OF    DIFFERENT    KINDS. 

TO    WHICH    IS   ADDED,      ^ 

THE   GEOMETRICAL    CONSTRUCTION   OF    A    GREAT    NUMBER 
OF    LINEAR    AND    PLANE    PROBLEMS, 

WITH  THE  METHOD  OF  RESOLVING  THE  SAME  NUMERICALLY, 


BY  THOMAS  SIMPSON,  F.  R.  S. 


FIRST   AMERICAN,  FROM   THE   EIGHTH    LONDON   EDITION. 


PHILADELPHIA : 


PRINTED  FOR  MATHEW  CAREY. 
SOLD    BY    C.    8c    A.    CONRAD    &    CO.;    BRADFORD    Sc    INSKEEP; 
HOPKINS     Sc    EARLE;     JOHNSON     8c     WARNER;     KIMBER    & 
CONRAD;    BIRCH    Sc    SMALL;    AND    EVERT    DUYCKINCK. 

PRINTED    BY    T.    ^  G.    PALMER. 


March  17,  1809. 


THE 

AUTHOR'S  PREFACE 

TO    THE 

SECOND  EDITION. 


The  motives  that  first  gave  birth  to  the  ensuing 
work,  were  not  so  much  any  extravagant  hopes  the 
author  could  form  to  himself  of  greatly  extending 
the  subject  by  the  addition  of  a  large  variety  of  new 
improvements  (though  the  reader  will  find  many 
things  here  that  are  nowhere  else  to  be  met  with), 
as  an  earnest  desire  to  see  a  subject  of  such  general 
importance  established  on  a  clear  and  rational  foun- 
dation, and  treated  as  a  science,  capable  of  demon- 
stration,  and  not  a  myterious  art,  as  some  authors, 
themselves,  have  thought  proper  to  term  it. 

How  well  the  design  has  been  executed,  must  be 
left  for  others  to  determine.  It  is  possible  that  the 
pains  here  taken,  to  reduce  the  fundamental  princi- 
pies,  as  well  as  the  more  difficult  parts  of  the  sub- 
ject to  a  demonstration,  may  be  looked  upon,  by 
some,  as  rather  tending  to  throw  new  difficulties  in 
the  way  of  a  learner,  than  to  the  facilitating  of  his 
progress.  In  order  to  gratify,  as  far  as  might  be, 
the  inclination  of  this  class  of  readers,  the  demon- 
strations  are  now  given  by  themselves,  in  the  man- 
ner  of  notes  (so  as  to  be  taken  or  omitted  at  plea- 
sure) ;  though  the  author  cannot  by  any  mieans  be 
induced  to  think  that  time  lost  to  a  learner  which 


PREFACE. 

is  taken  up  in  comprehending  the  grounds  whereon 
he  is  to  raise  his  superstructure :  his  progress  may, 
indeed,  at  first,  be  a  little  retarded;  but  the  real 
knowledge  he  thence  acquires  will  abundantly  com- 
pensate his  trouble,  and  enable  him  to  proceed,  af- 
terwards, with  certainty  and  success,  in  matters  of 
greater  difFculty,  where  audiors,  and  their  rules,  can 
yield  him  no  assistance,  and  he  has  nothing  to  de- 
pend upon  but  his  own  observation  and  judgment. 

This  second  edition  has  many  advantages  over 
the  former,  as  well  with  respect  to  a  number  of  new 
subjects  and  improvements,  interspersed  throughout 
the  whole,  as  in  the  order  and  disposition  of  the  ele- 
mentary parts  :  in  which  particular  regard  has  been 
had  to  the  capacities  of  young  beginners.  The 
work,  as  it  now  stands,  will,  the  author  flatters  him- 
self, be  found  equally  plain  and  comprehensive,  so 
as  to  answer,  alike,  the  purpose  of  the  lower  and 
of  the  more  experienced  class  of  readers. 


P.  S.  The  great  reputation  of  Mx.  Simpson's 
Treatise  o/'Algebra,  and  the  favourable  recep^ 
tion  it  has  universally  met  with  since  its  first  publi- 
cation^  and  which  testifies  it  to  be  the  best  elementary 
work  upon  the  subject^  has  induced  the  proprietor  to 
have  this  edition  carefully  revised  and  corrected 
throughout  by  a  very  eminent  mathematician:  he 
therefore  trusts  it  will  be  found  as  worthy  the  appro- 
bation of  the  public  as  any  former  edition^  or  as  if 
revised  by  the  author  himself 


THE  CONTENTS. 


SECTION  L 

NOTATION 

SECTION  IL 

Page  1 

ADDITION 

SECTION  III. 

8 

SUBTRACTION 

SECTION  IV. 

11 

MULTIPLICATION 

13 

SECTION  V. 

DIVISION 

SECTION  VI. 

28 

INVOLUTION 

SECTION  vn. 

36 

EVOLUTION 

- 

42 

SECTION  VIIL 
THE  REDUCTION  OF  FRACTIONAL  AND  RA- 

DICAL  QUANTITIES  -  45 

SECTION  IX. 
OF  EQUATIONS  -  -  57 

1 .  The  reduction  of  single  Equations  ibid. 

2.  The  extermination  of  unknown  Quantities,  or  the 
reduction  of  two  or  more  equations  to  a  single  one       63 

SECTION  X. 
OF  ARITHMETICAL  AND  GEOMETRICAL 

PROPORTIONS  -  69 

SECTION  XL 
THE  SOLUTION  OF  ARITHMETICAL  PROB- 
LEMS -  "7^ 
SECTION  XII. 
THE  RESOLUTION  OF  EQUATIONS  OF    SE- 

VERAL  DIMENSIONS  -  131 

1.  Of  the  origin  and  composition  of  equations  ibid. 

2.  How  to  know  whether  some,  or  all  the  roots  of  an 

Equation  be  rational,  and,  if  so,  what  they  are  1 34 


CONTENTS. 

Page 

3.  Another  way  of  discovering  the  same  thing,  by- 

means  of  Sir  Isaac  Newton's  method  of  divisors ; 
with  the  grounds  and  explanation  of  that  method      136 

4.  Of  the  solution  of  cubic  Equations  according  to 

Cardan    '  -  143 

5.  The  same  method  extended  toother,  higher  Equa- 

tions -  145 

6.  Of  the  solution  of  biquadratic. Equations  according 

to  Des  Cartes  -  148 

7.  The  solution  of  biquadratics  by  a  new  method,  with- 

out the  trouble  of  exterminating  the  second  term     150 

8.  Cases  of  biquadratic  Equations  that  may  be  re- 

duced to  quadratic  ones  -  153 

9.  The  resolution  of  literal  Equations,  wherein  the 

given,  and  the  unknown  quantity  are  alike  af- 
fected -  156 

10.  The  resolution  of  Equations  by  the  common  me- 

thod of  converging  series  -  158 

11.  Another  way,  more  exact  162 

12.  A  third  method         ^  -  170 

13.  The  method  of  converging  series  extended  to  surd 

Equations  -  -  174 

14.  A  method  of  solving  high  Equations,  when  two,  or 

more  unknown  quantities  are  concerned  in  each       177 
SECTION  XIII. 
OF  INDETERMINATE  PROBLEMS  180 

SECTION   XIV. 
THE    INVESTIGATION    OF    THE    SUMS    OF 

POWERS         -  -201 

SECTION  XV. 
OF  FIGURATE  NUMBERS  -  213 

1 .  The  Sums  of  Series,  consisting  of  the  reciprocals  of 

figurate  numbers,  with  others  of  the  like  nature       2 1 5 
*2,  The  sums  of  compound  Progressions,  arising  from 
a  series  of  powers  drawn  into  the  terms  of  a  geo- 
metrical progression  219 

3.  The  combinations  of  Quantities      ^        -  225 

4.  A  demonstration  of  Sir  Isaac  Newton's  Binomial 

theorem  -  -  227 

SECTION   XVI. 
OF  INTEREST  AND  ANNUITIES         -  229 

1.  Annuities  and  Pensions  in  Arrear,  computed   at 

simple  interest  -  231 


CONTENTS. 

Page 

2.  The  investigation  of  Theorems  for  the  solution  of  the 

various  cases  in  compound  interest  and  annuities    234 
SECTION  XVII. 
OF  PLANE  TRIGONOMETRY  -  241 

SECTION  XVIII. 
THE    APPLICATION    OF   ALGEBRA  TO   THE 
SOLUTION     OF     GEOMETRICAL     PROB- 
LEMS  .  -  254 

1 .  An  easy  way  of  constructing,  or  finding  the  roots  of 

a  quadratic  equation,  geometrically  267 

2.  A  demonstration  why  a  problem  is  impossible  when 

the  square  root  of  a  negative  quantity  is  concerned     272 

3.  A  method  for  discovering  whether  the  root  of  a  ra- 

dical quantity  can  be  extracted  -  284 

4.  The  manner  of  taking  away  radical  quantities  from 

the  denominator  of  a  fraction,  and  transferring 
them  to  the  numerator  -  288 

5.  A  method  for  determining  the  roots  of  certain  high 

Equations,  by  means  of  the  secti(Ai  of  an  angle         30 1 
AN  APPENDIX, 
Containing  the  geometrical  construction  of  a  large  varie- 
ty of  linear  and  plane  Problems;  with  the  man- 
ner of  resolving  the  same  numerically  3 1 5 


TREATISE 

OF 

A  L  G  E  B  R  A- 


SECTION  I. 

Of  Notation. 

XjlLGEBRA  is  that  science  which  teaches,  in  a  general 
manner,  the  relation  and  comparison  of  abstract  quanti- 
ties :  by  means  whereof  silch  questions  are  resolved  whose 
solutions  would  be  sought  in  vain  from  common  arithme- 
tic. 

In  algebra,  otherwise  called  specious  arithmetic^  num- 
bers are  not  expressed  as  in  the  common  notation,  but 
every  quantity,  whether  given  or  required,  is  commonly 
represented  by  sqme  letter  of  the  alphabet;  the  given  ones, 
for  distinction  sake,  being  usually  denoted  by  the  initial 
letters  a,  ^,  r,  d^  &c. ;  and  the  unknown  or  required  ones 
by  the  final  letters  7/,  w,  x,  ?/,  &c.  There  are,  moreover, 
in  algebra,  certain  signs  or  notes  made  use  of  to  show  the 
relation  and  dependence  of  quantities  one  upon  another, 
whose  signification  the  learner  ought,  first  of  all,  to  be 
made  acquainted  with. 

The  sign  4-  signijies  that  the  quantity   to  which  it  is 
prefixed  is  to  be  added.       Thus  a  +  b  shows    that  the 

B 


2  Of  Notation. 

number  re^presented  by  b  is  to  be  added  to  that  repre^ 
sented  by  «,  and  expresses  the  sum  of  those  numbers ;' 
so  that  if  a  was  5,  and  b  3,  then  would  a  +  b  ht  5  +3^ 
or  8.  In  like  manner,  a  +  b  +  c  denotes  the  number 
arising  by  adding  all  the  three  numbers  a,  b^  and  c  toge- 
ther. 

Note.  A  quantity  which  has  no  prefixed  sign  (as  the 
leading  quantity  a  in  the  above  examples)  is  always  under- 
stood to  have  the  sign  -f-  before  it :  so  that  a  signifies  the 
same  as  +  ^  ;  and  a-^-b  the  same  as  +a  +  b. 

The  sign  —  signifies  that  the  quantity  which  it  precedes 
is  to  be  subtracted.  Thus  a  —  b  shows  that  the  quan- 
tity represented  by  b  is  to  be  subtracted  from  that  repre- 
sented by  a,  and  expresses  the  difference  of  a  and  bi 
so  that  if  a  was  5,  and  b  3,  then  would  a  —  b  be  5  —  3, 
or  2.  In  like  manner,  a  +  b  —  c  —  d  represents  the 
quantity  which  arises  by  taking  the  numbers  c  and  d 
from  the  sum  of  the  other  two  numbers  a  and  b :  as, 
if  a  was  7,  <^  6,  c  5,  and  d  3,  then  would  a  +  b  —  c  —  d  be 
7  +  6  —  5  —  3,  or  5. 

The  notes  +  and  —  are  usually  expressed  by  the  words 
plus  (or  7nore)^  and  minus  (or  less).  Thus,  we  read  a  +  b^ 
a  plus  b  \  and  a  —  b^a  minus  b. 

Moreover,  those  quantities  to  which  the  sign  +  is  pre- 
fixed are  called  positive  (or  ajfirmative)  \  and  those  to 
which  the  sign  —  is  prefixed,  negative. 

The  sign  X  signifies  that  the  quantities  betzveen  which 
it  stands  are  to  be  ynultiplied  together.  Thus  axb  denotes 
that  the  quantity  a  is  to  be  multiplied  by  the  quantity 
b^  and  expresses  the  product  of  the  quantities  so  multi- 
plied ;  and  ax  b  X  c  expresses  the  product  arising  by 
multiplying  the  quantities  a,  b^  and  c,  continually  to- 
gether :  thus,  likewise,  a-^-b  X  c  denotes  the  product  of 
the  compound  quantity  a  +  b  by  the  simple  quantity 
c  ;  and  a  +  b  +  c  X  a  —  b  +c  X  a  +  c  represents  the 
product  which  arises  by  multiplying  the  three  com- 
pound quantities  a  +  b  +  c^  a  —  b  +  c^  and  a  +  c  con- 
tinually together ;   so  that  if  a  was  5,  b  4,  and  c  3,  then 


OfNotatmu 


would  a  +  /^+cX«  —  ^H-cx«  +  cbe  12x4x8, 
which  is  384. 

But  when  quantities  denoted  by  single  letters  are 
to  be  multiplied  together,  the  sign  x  is  generally- 
omitted,  or  only  understood ;  and  so  ab  is  made  to 
signify  the  same  as  aXb  \  and  ahc  the  same  2&  axh 
X  c. 

It  is  likewise  to  be  observed,  that  when  a  quantity  is 
to  be  multiplied  by  itself,  or  raised  to  any  power,  the 
usual  method  of  notation  is  to  draw  a  line  over  the  given 
quantity,  and  at  the  end  thereof  place  the  exponent 
of  the  power.  Thus  a  +  b\^  denotes  the  same  as  a -f- ^ 
X  a  -f  ^,  viz.  the  second  power  (or  square)  oi  a  +  b  con- 
sidered as  one  quantity  :  thus,  also,  ab  +  be']  ^  denotes  the 
same  as  ab  +  bc  X  cib -{-be  x  ab  +  bc^  viz.  the  third  power 
(or  cube)  of  the  quantity  ab  +  be. 

But,  in  expressing  the  powers  of  quantities  repre- 
sented by  single  letters,  the  line  over  the  top  is  com- 
monly omitted;  and  so  a^  comes  to  signify  the  same 
as  au  or  ax  a,  and  b  ^  the  same  as  bbb  or  b  X  b  X  b  : 
whence  also  it  appears  that  a^  b^  will  signify  the  same 
as  aabbb ;  and  a^e^  the  same  as  aaaaaee ;  and  so  of 
others. 

The  note  .  (or  a  full  point),  and  the  word  into^  are 
likewise  used  instead  of  x,  or  as  marks  of  multiplica- 
tion. 

Thus  a  +  b  .  a  -f-  c  and  a-^b  into  a  +  e^  both  signify 
the  same  thing  as  a  +  b  x  a  +  c^  namely,  the  product  of 
a  +  bhy  a  +  c. 

The  sign  -r-  is  used  to  signify  that  the  quantity  pre^ 
ceding  it  is  to  be  divided  by  the  quantity  which  comes 
after  it :  thus  c  -—  ^  signifies  that  c  is  to  be  divided  by 
b ;  and  a  +  b-rra  —  c,  that  a-^b  is  to  be  divided  by 
a  —  c. 

Also  the  mark  )  is  sometimes  used  as  a  note  of  divi- 
sion ;  thus,  a  +  b^  ab^  denotes  that  the  quantity  ab  is 
to  be  divided  by  the  quantity  a-^-b  \  and  so  of  others. 
jBut  the  division  of   algebraic  quantities  is  most  com- 


4  Of  Notation. 

monly  expressed  by  writing  down  the  divisor  under  the 

dividend  with  a  line  between  them,  in  the  manner  of 

c 
a    vulgar    fraction.        Thus    ~-j-  represents  the  quantity 

arising  by  dividing  c  by  ^  ;  and denotes  the  quan- 
tity arising  by  dividing  a  +  bhy  a  —  c.  Quantities  thus 
expressed  are  called  algebraic  fractions ;  whereof  the 
upper  part  is  called  the  numerator,  and  the  lower  the  de- 
nominator, as  in  vulgar  fractions. 

The  sign  V  is  used  to  express  the  square  root  of 
any  quantity  to  which  it  is  prefixed:  thus  V25  signi- 
fies the  square  root  of  25  (which  is  5,  because  5  X  5  is 
25)  :  thus  also  \^cib  denotes  the  square  root  of  ab  \    and 

W — Z— «    denotes    the    square    root  of — ,    or   of 

^      a  a 

the    quantity  which    arises    by  dividing   ab  +  be  by  d  -, 

but  — — ^t —    (because    the    line    which    separates    the 
d 

numerator  from  the  denominator  is  drawn  below  V  ) 
signifies  that  the  square  root  of  ab  +  be  is  to  be  first 
taken,  and  afterwards  divided  by  d :   so  that  if  a  was  2, 

7  n        A        A  ^  c^    4\.  ij  ^'oT+bc  ,      V36         6  . 

6)  6,  c  4,  and  a  9,  then  would  ■ be or  — ; 

'    _J d  9         9 

but  y  2— Z —  is  s^—-i  or  ^4,  which  is  2. 

The  same  mark  -^z,  with  a  figure  over  it,  is  also  used 
to  express  the  cube,  or  biquadratic  root,  &fc.  of  any 
quantity :  thus  v^64  represents  the  cube  root  of 
64  (which  is  4,  because  4x4x4  is  64),  and  Vab  +  ed 
the  cube  root  of  ab  -{-  cd  ;  also  Vl6  denotes  the 
biquadratic  root  of  16  (which  is  2,  because  2  x  2  X  2  x 
2  is  16);  and  Vab  +  cd  denotes  the  biquadratic  root 
of  ab-^cd;  and  so  of  others.  Quantities  thus  ex- 
pressed are  called  radical  quantities,  or  surds  ;    where- 


Of  Notation.  5 

of  those  consisting  of  one  term  only,  as  Va  and  \^aby 
are  called  simple  surds ;  and  those  consisting  of  several 
terms,  or  members,  as  Va^  — b^  and  Va^  —  b^  +  bc^  com- 
pound  surds. 

Besides  this  way  of  expressing  radical  quantities, 
(which  is  chiefly  followed)  there  are  other  methods 
made  use  of  by  different  authors  ;  but  the  most  com- 
modious of  all,  and  best  suited  to  practice,  is  that,  where 
the  root  is  designed  by  a  vulgar  fraction,  placed  at  the 
end  of  a  line  drawn  over  the  quantity  given.  Ac- 
cording to  this  notation,  the  square  root  is  designed  by 
the  fraction  |,  the  cube  root  by  |,  and  the  biquadratic 
root  by  l,  &?c.  Thus  a  p  expresses  the  same  thing  with 
Va^  viz.  the  square  root  of  a ;  and  a^  +  ab'\z  the 
same  as  Va^  +  ab^  that  is,  the  cube  root  of  a^  ^ab: 
also  a\i  denotes  the  square  of  the  cube  root  of  a ;  and 
<2-f  z")*  the  seventh  power  of  the  biquadratic  root  of 
a-\-z',  and  so  of  others.  But  it  is  to  be  observed,  that, 
when  the  root  of  a  quantity  represented  by  a  single 
letter  is  to  be  expressed,  the  line  over  it  may  be  ne- 
glected ;  and  so  a^  will  signify  the  same  as  a"]^,  and  b\ 
the  same  as  ^  (3  or  Vb.  The  number,  or  fraction,  by 
which  the  power,  or  root  of  any  quantity  is  thus  designed 
is  called  its  index,  or  exponent. 

The  mark  r^  {called  the  sign  of  equality)  is  used  to 
signify  that  the  quantities  standing  on  each  side  of  it  are 
equal.  Thus  2  -f  3  =  5,  shows  that  2  more  3  is  equal  to 
5  ;  and  x=.a-^bj  shows  that  x  is  equal  to  the  difference 
of  a  and  b. 

The  note  : :  signifies  that  the  quantities  between  which 
it  stands  are  proportional :  as,  a  :  b  :  i  c  '.  d^  denotes 
that  a  is  in  the  same  proportion  to  ^,  as  c  is  to  ^;  or  that 
if  a  be  twice,  thrice,  or  four  times,  ^c.  as  great  as  b^ 
then  accordingly  c  is  twice,  thrice,  or  four  times,  i^c.  as 
great  as  d. 


t>  Of  Notation* 

To  what  has  been  thus  far  laid  down  on  the  signifi- 
cation of  the  signs  and  characters  used  in  the  algebraic 
notation,  we  may  add  what  follows,  which  is  equally  ne- 
cessary to  be  understood. 

When  any  quantity  is  to  be  taken  more  than  once, 
the  number  is  to  be  prefixed,  which  shows  how  many 
times  it  is  to  be  taken :  thus  5a  denotes  that  the  quan- 
tity a  is  to  be  taken  five  times ;  and  Zhc  stands  for  three 
times  hc^  or  the  quantity  which  arises  by  multiplying  he 
by  3  :  also  7Va^  -f  h^  signifies  that  >/a^  ■^h'^  is  to  be 
taken  seven  times ;  and  so  of  others. 

The  numbers  thus  prefixed  are  called  coefficients  ^ 
and  that  quantity  which  stands  without  a  coefficient  is 
always  understood  to  have  a  unit  prefixed,  or  to  be  taken 
once,  and  no  more. 

Those  quantities  are  said  to  be  like  that  are  expressed' 
by  the  same  letters  under  the  same  powers,  or  which 
differ  only  in  their  coefficients :    thus  3^c,  5hc^  and  ^hc 
are  like  quantities  ;  and  the  same  is  to  be  understood  of 

the  radicals  2v/  and  *ts}         .      But  unlike  quan- 

^     a  ^     a 

tities  are  those  which  are  expressed  by  different  letters, 

or  by  the  same  letters  tinder  different  powers :  thus  2aby 

2abc^  5ab^^  and  oba^  are  all  unlike* 

When  a  quantity  is  expressed  by  a  single  letter,  or  b\' 
s'everal  single  letters  joined  together  in  multiplication 
without  any  sign  between  them,  as  a,  or  2ab^  it  is  called 
a  simple  quantity. 

But  that  quantity  which  consists  of  two  or  more  such 
simple  quantities,  connected  by  the  signs  +  or  — ,  is  called 
a  compound  quantity  :  thus  a  —  2ab  -f-  5abc  is  a  compound 
quantity ;  whereof  the  simple  quantities  a,  2<3^,  and  5abc 
are  called  the  terms  or  members. 

The  letters  by  which  any  simple  quantity  is  expressed 
may  be  ranged  according  to  any  order  at  pleasure,  and 
yet  the  signification  continue  the  same :  thus  ab  may 
be  written  ba  ;  for  ab  denotes  the  product  of  a  by  ^,  and 
ba  the  product  of  ^  by  a ;  but  it  is  well  known,  that 
when  two  numbers  are  to  be  multiplied  together,  it 
matters  not  which  of   them  is  made  the  multiplicand. 


Of  Notation*  t 

nor  which  the  multiplier,  the  product,  either  way, 
coming  out  the  same.  In  like  manner  it  will  appear 
that  abc^  acb^  bac^  bca^  cab^  and  cba^  all  express  the  same 
thing,  and  may  be  used  indifferently  for  each  other,  as 
will  l3e  demonstrated  further  on ;  but  it  will  be  some- 
times found  convenient,  in  long  operations,  to  place  the 
several  letters  according  to  the  order  which  they  have 
in  the  alphabet. 

Likewise  the  several  members  or  terms  of  which  any 
quantity  is  composed  may  be  disposed  according  to  any 
order  at  pleasure,  and  yet  the  signification  be  nowise 
affected  thereby.  Thus  a  —  2ab  -j-  Sa^b  may  be  written 
a  +  5a^b  —  2ab,  or  —  2ab  +  a  +  5a^b,  kc. ;  for  all  these 
represent  the  same  thing,  viz.  the  quantity  which  remains, 
when,  from  the  sum  oi  a  and  5a^b^  the  quantity  2ab  is  de- 
ducted. 

Here  follow  some  examples  wherein  the  several  forms 
of  notation  hitherto  explained  are  promiscuously  con- 
cerned, and  where  the  signification  of  each  is  expressed 
in  numbers. 

Suppose  a  =  6,  ^  =  5,  and  c  =  4  ;  then  will 
a2  +sab  —  c2  =  36  +  90  —  16  =  110, 
2a^  —  3a^b  +c^  =  432  ~  540  +  64  =  —  44, 
^2  xa  +  b  —  2abc=S6x  H  —240=  156, 

a^                    216 
+  c2  = +  16  =  12  +  16  =  28, 

V2ac  +  c^(or  2ac  +  c^  H)  =  V64  =  8  (for  8  X  8  =  64), 

2hc  ^      40 

a2  ^S/l)2  ^^^  _  36  —  1  ___  35  _ 

2a'--\/b^  +ac  ""l2  — 7^T""    ' 

V^ 2  ^ac^ x/2ac  +  c^=  1+8  =  9, 

^b^  —  ac  +  V2ac  +  c2*  =  V25  —  24  +  8  =  3. 

This  method  of  explaining  the  signification  of  quan- 
tities I  have  found  to  be  of  good  use  to  young  begin- 
ners ;  and  would  recommend  it  to  such  as  are  desirous 
of  making  proficiency  in  the  subject,  to  get  a  clear  idea 


8  Of  Addition. 

of  what  has  been  thus  far  delivered,  before  they  proceed 
farther. 


SECTION  IL 

Of  Addition. 


ADDITION,  in  algebra,  is  performed  by  connecting 
the  quantities  by  their  proper  signs,  and  joining  into  one 
sum  such  as  can  be  united :  for  the  more  ready  effecting  of 
which,  observe  the  following  rules. 

1°.  If  in  the  quantities  to  be  added^  there  be  terms  that 
are  like^  and  have  all  the  same  sig^n^  add  the  coefficients  of 
those  terms  together^  and  to  their  sum  adjoin  the  letters 
common  to  each  term^  prefixing  the  common  sign. 

Thus  Sa      And  5a  +  7b     Also         5a — 7b 

added  to       3a      added  to    7a  -f-  3^     added  to  7a  —  3<^ 


makes            Sa.^makes      12a  +  10^.  makes  12a  —  10b. 

Hence         {''^Vab  +  7Vbc          *     ,  ^,      f^^  Sd 

I  ^    , —            ,—         And  the  j  —  —         — 

likewise     <^Wab+  2Vb^         sum  of     1   ^  ^ 

the  sum  of  ^\S>Vab  -f  WTc                         l^A  _        ^ 
will  be         11  Vab+ 1 8  VTc 


a 


.„  ,       7b  lOd 

will  be    —        —      — 


The  reasons  on  which  the  preceding  operation^  are 
grounded  will  readily  appear  by  reflecting  a  little  on  the 
nature  and  signification  of  the  quantities  to  be  added : 
for,  with  regard  to  the  first  example,  where  Sa  is  to  be 
added  to  5a,  it  is  plain  that  three  times  any  quantity 
whatever,  added  to  five  times  the  same  quantity,  must 
make  eight  times  that  quantity :  therefore  3a,  or  thi*ee 
times  the  quantity  denoted  by  a,  being  added  to  5a^  or 
five  times  the  same  quantity,  the  sum  must  consequently 


Of  Addition.  9 

2"*.  JV/ieTij  271  the  quantities  to  be  added^  there  are  like 
terms ^  whereof  some  are  affrmative  and  others  negative^ 
add  together  the  affirmative  terms  (if  there  be  more  than 
one^^  and  do  the  same  by  the  negative  ones;  then  take  the 
difference  of  the  two  sums  {not  regarding  the  signs)  by 
subtracting  the  coefficient  of  the  lesser  from  that  of  the 
greater^  and  adjoin  the  letters  common  to  each;  to  which 
difference  prefix  the  sign  of  the  greater* 

Examples  of  this  rule  may  be  as  follows. 


I2a  —  5b 

2. 

—  Sab +5  be 

—  Sa  +2b 

^7ab  —  9ac 

Sum       9a  —  3^ 

Sum       4«Z> — 4^bc 

6ab  +  12bc —   Scd 

4. 

5Vab  —  7Vbc  +    Sd 

—  Tab—   9bc+    3cd 

SVab  +  SVbc —  12d 

—  2ab^   5bc+  12cd 

7\'ab  4-  3v/;c  4-    9d 

Sum  —  2,ab —  2bc  +    7cd       Sum  15V  ab  +  4V/^c  +    5d 

5.  1 2abc  —  1  ^abd  +  25acd—  72bcd 

1  ^abc  +  1 2abd  -f  20acd'^  1  Sbcd 

—  1  Sabc  —  26abd—  1 5acd  +12bcd 

82abc+  ISabd —  1  Oacd —  1 6bcd 

Sum  47abc — 12abd  -f-  20acd —  94:bcd 


make  8a,  or  eight  times  that  quantity.  From  whence,  as 
the  sum  of  any  two  quantities  is  equal  to  the  sum  of  all 
their  parts,  the  reason  of  the  second  case,  or  example,  is 
likewise  obvious.  But  as  to  the  third,  where  the  given 
quantities  are  5a  —  7b  and  7a  —  Sbj  we  are  to  consider, 
that,  if  the  two  quantities  to  be  added  together  had  been 
exactly  5a  and  7a  (which  are  the  two  leading  terms),  the 
sum  would  then  have  been  just  12a;  but,  since  the  for- 
mer quantity  wants  7b  of  5a,  and  the  latter  Sb  of  7a^ 
their  sum  must,  it  is  evident,  want  both  7b  and  3^  of  12a ; 
and  therefore  be  equal  to  12a — 106,  that  is,  equal  to 
what  remains,  when  the  sum  of  the  defects  is  deducted. 
And,  by  the  very  same  way  of  arguing,  it  is  easy  to  con- 
ceive that  the  sum  which  arises  by  adding  any  number 


10  Of  Addition. 

6.          oa '^cc  ^    i^^^q    \^E^^ 

b         a  ^  a          ^       a 

ha  ^  a          ^       a 

^         13a  .  4cc  ^    fbc       ^    lab  A- 

Sum  -r-  + S\} 3\/ — JL 

h              n  ^  n              ^         n 


CC 

a 


tn  the  last  example,  and  all  others  where  fractional 
and  radical  quantities  are  concerned,  every  such  Quan- 
tity, exclusive  of  its  coefficient,  is  to  be  treated  in 
all  respects  like  a  simple  quantity  expressed  by  a  singlc 
letter. 

3°.  When^  in  the  quantities  to  be  added^  there  are  terms 
without  others  like  to  them^  write  them  down  with  their 
proper  signs* 

Thus         a  +  2<^  And         aa-^-bh 

added  to  3c  -f-  <f         ,  added  to    a-{-b 

makes       a  +  2<^  +  3c  +  <^       makes      aa  -^  bb  +  a  -j-  b 
Here  follow  a  few  examples  for  the  learner's  exercise, 
wherein  all  the  three  foregoing  rules  take  place  promis- 
cuously. 

1.  2aa  +  3ab+    8cc  +  d^ 

5aa  —  7cd)  +    5cc — d^ 

—  2aa  -f  4ab  +    3cc  -f  30 

Sum       5aa  ^         +  16cc  +  J^  —  ^s  ^30. 


of  quantities  together,  will  be  equal  to  the  sum  of  all  the 
affirmative  terms  diminished  by  the  sum  of  all  the  ne- 
gative ones  (considered  independent  of  their  signs)  :  from 
whence  the  reason  of  the  second  general  rule  is  apparent. 
As  to  the  case  where  the  quantities  are  unlike,  it  is  plain 
that  such  quantities  cannot  be  united  into  one,  or  other- 
wise added  than  by  their  signs  ;  thus,  for  example,  let  a 
be  supposed  to  represent  a  crown,  and  b  a  shilling  ;  then 
the  sum  of  a  and  b  can  be  neither  2a  nor  2^,  that  is,  nei- 
ther two  crowns  nor  two  shillings,  but  one  crown  plus  one 
shilling,  Qv  a  +  b. 


Of  Subtraction^  .  1 1 

2.  BVqx  »—    ^Vaa  — XX  +  1 2\^aa  -f-  4:XX 

^Vax  +  \5Vaa  —  xx —   ^\^aa  +  4^xx 

^S/ax —    7\^aa  —  xx  -f-  10V«c/  4-  ^^xx 

Sum  19  Vox  ^  +14V««+4.rI' 

4c3  _    Hh^  ^5ah'\'  100 

20g<^  +  16^2  —   ^c  —  80 ___^ 

Sum  13rt2  +  22«^  +  3Z>3  +  ^3  _  ^.  3  ^20  —  bi:7 


SECTION  III. 

Of  Subtraction. 

SUBTRACTION,  in  algebra,  is  performed  by  change 
ing  all  the  signs  of  the  subtrahend  {or  conceiving  them  to 
be  changed),  and  then  connecting  the  quantities,  as  in  addi- 
tion. 

Ex.  1.  From  8«  +  5b  Ex.  2.  From  85  +  5b 

take     5a  +  ^b  take     5a  —  Zb 

Rem.  3a  +  2b.  Rem.   ':ia  +  8b. 

Ex.  3.  From  8a  —  5b  Ex.  4.  From  Sa — 5b 

take     5a  -f  3/?  take    5a —  3b 

Rem.  3a— 8^.  Rem.  3a  —  2b. 

In  the  second  example,  conceiving  the  signs  of  the 
subtrahend  to  be  changed  |o  their  contrary,  that  of 
3b  becomes  -f ;  and  so  the  signs  of  3b  and  5b  being 
alike,  the  coefficients  3  and  5  are  to  be  added  together, 
by  case  1  of  addition.  The  same  thing  happens  in  the 
third  example  ;  since  the  sign  of  3b,  when  changed,  is  — , 
and  therefore  the  same  with  that  of  5b.  But,  in  the  fourth 
example,  the  signs  o£  3b  and  5b,  after  that  of  3b  is  chang- 
ed, being  unlike,  the  difference  of  the  coefficients  must  be 
taken,  conformable  to  case  2  in  addition. 


12  Of  Subtraction, 

Other  examples  in  subtraction  may  be  as  follows : 
From    20ax  +  5bc  —  7aa  From      7Vax  +  9>/b  y 

take       I'^ax  —  Zhc  —  Saa  take    — 5Vflx  +  12v'^i!/ 

Rem.       ^ax  +  %bc  —  %aa.  Rem.    12\  «^—  Z*^Vy. 

From      6  V<2«  —  ;cx  +  loVo^  —  :v^ —  7v  22 


take         9\/«<:/  —  xX'\-\  Sy/a^  —  x"^  —  ^si— 

^  c 


^  „ s ,--  faa 

Rem.  —  SVaa  —  xx  +25Va^  ~x^  +  2>J— 

From  Ta^^^-^eJ^  +  d 
c  ^   c 

1             c>   .  Sa  fax  ,  , 

take        a^  -\ V  —  +  ^ 


Rem.  6a^>^  —  +  7sr^  +  d—b. 
c  ^  c 

In  this  last  example,  the  quantity  cf  in  the  subtrahend 
being  without  a  coefficient,  a  unit  is  to  be  understood  ;  for 
la^  and  a^  mean  the  same  thing.  The  like  is  to  be  ob- 
served in  all  other  similar  cases. 


The  grounds  of  the  general  rule  for  the  subtraction 
of  algebraic  quantities  may  be  explained  thus:  let  it  be 
here  required  to  subtract  5a  —  Sb  from  Sa  +  5b  (as  in  ex. 
2).  It  is  plain,  in  the  first  place,  that  if  the  affirma- 
tive part  5a  were  alone  to  be  subtracted,  the  remainder 
would  then  be  8a  +  5b  <-^  5a;  but,  as  the  quantity  actu- 
ally proposed  to  be  subtracted  is  less  than  5«,  by  3^,  too 
much  has  been  taken  away  by  3^ ;  and  therefore  the  true 
remainder  will  be  greater  than  8a  +  5b  —  5«,  by  Sb  ;  and 
so  will  be  truly  expressed  by  8a  +  5b  —  5a-j-Sb:  where- 
in the  signs  of  the  two  last  terms  are  both  contrary  to  what 
they  were  given  in  the  subtrahend ;  and  where  the  whole, 
by  uniting  the  like  terms,  is  reduced  to  3a  -{-  8b^  as  in  the 
example. 


Of  Multiplication.  f^ 

SECTION  IV. 

Of  Multiplication* 

BEFORE  I  proceed  to  lay  down  the  necessary  rules 
for  multiplying  quantities  one  by  another,  it  may  be  pro- 
per to  premise  the  following  particulars,  in  order  to  give 
the  learner  a  clear  idea  of  the  reason  and  certainty  of  such 
rules. 

Firsts  then^  it  is  to  he  observed^  that  when  several 
quantities  are  to  be  multiplied  continually  together^  the  re- 
sult^ or  product^  will  come  out  exactly  the  same^  midtiply 
them  according  to  what  order  you  will.  Thus  a  X  b  x  (\ 
a  X  c  X  b^  b  X  c  X  a^  &fc.  have  all  the  same  value,  and 
may  be  used  indifferently :  to  illustrate  which  we  maj'^ 
suppose  c  =  2,  ^  =  3,  and  c  =  4;  then  will  a  X  b  x  c-=z 
2X3X4  =  24;  axcx^  =  2x4x3  =  24;  and  bxc 
X<^  =  3X4X2  =  24. 

Secondly.  If  any  number  of  quantities  be  multiplied 
continually  together^  and  any  other  number  of  quantities 
he  also  multiplied  continually  together^  and  then  the  two 
products  one  into  the  other ^  the  quantity  thence  arising  will 
be  equal  to  the  quantity  that  arises  by  multiplying  all  the 
proposed  quantities  continually  together.  Thus  will  abc  X 
de  z=  a  X  b  X  c  X  d  X  e  \  so  that,  if  a  were  =  2,  ^  =  3,  c  =  4, 
d=5^e=z6^  then  would  abc  X  de  =  24x  S0=  720,  and 
flX<^Xcx^Xe  =  2x3x4x5x6  =  720.  The  ge- 
neral demonstration  of  these  observations  is  given  below 
in  the  notes. 


The  following  demonstrations  depend  on  this  prin- 
ciple, that  if  two  quantities^  whereof  the  one  is  n  times  as 
great  as  the  other  (n  being  any  number  at  pleasure)^  be 
multiplied  by  one  and  the  same  quantity,,  the  product,,  in 
the  one  case,,  will  also  be  n  times  as  great  as  in  the  other. 
The  greater  quantity  may  be  conceived  to  be  divided 
into  n  parts,  equal,  each,  to  the  lesser  quantity;  and 
the  product  of  each  part  (by  the  given  multiplier)  will 


14  OJ  Multiplication. 

The  multiplication  of  algebraic  quantities  may  be  con- 
sidered in  the  seven  following  cases. 


be  equal  to  that  of  the  said  lesser  quantity ;  therefore  the 
sum  of  the  products  of  all  the  parts  which  make  up  the 
whole  greater  product,  must  necessarily  be  n  times  as 
great  as  the  lesser  product,  or  the  product  of  one  single 
part,  alone. 

This  being  premised,  it  will  readily  appear,  in  the  first 
place,  that  b  x  ci  and  a  x  b  are  equal  to  each  other : 
for,  b  X  CL  being  b  times  as  great  as  1  x  «  {because  the 
niultiplicajid  is  b  times  as  great^  it  must  therefore  be  equal 
to  1  X  ci  (or  a)  repeated  b  times,  that  is,  equal  to  ax  b^  by 
the  definition  of  multiplication* 

In  the  same  manner,  the  equality  of  all  the  variations, 
or  products,  abc^  bac^  acb^  cab^  bca^  cba  (where  the  num- 
ber of  factors  is  3)  may  be  inferred :  for  those  that  have 
the  last  factors  the  same  (xvhich  I  call  of  the  same  class^ 
are  manifestly  equal,  being  produced  of  equal  quantities 
multiplied  by  the  same  quantity :  and,  to  be  satisfied 
that  those  of  different  classes^  as  abc  and  acb^  are  like- 
wise equal,  we  need  only  consider,  that,  since  ac  x  b 
is  c  times  as  great  as  «  X  <^  (because  the  multiplicand  is  c 
times  as  great)  it  must  therefore  be  equal  Xo  aX  b  taken  c 
times,  that  is,  equal  to  a  x  ^  X  c,  by  the  definition  of  mid- 
tiplication. 

Universally.  If'  all  the  products,  when  the  number 
of  factors  is  7z,  be  equal,  all  the  products,  when  the 
number  of  factors  is  ?2  +  1,  will  likewise  be  equal : 
for  those  of  the  same  class  are  equal,  being  produced 
of  equal  quantities  multiplied  by  the  same  quantity: 
and  to  show  that  those  of  different  classes  are  equal 
also,  we  need  only  take  two  products  which  differ  in 
their  tv/o  last  factors,  and  have  all  the  preceding  ones 
acGording  to  the  same  order,  and  prove  them  to  be 
equal.  These  two  factors  we  will  suppose  to  be  repre- 
sented by  r  and  5,  and  the  product  of  all  the  preceding 
ones  by  /; ;  then  the  two  products  themselves  will  be 
represented  by  prs  and  psr^  which  are  eqTial,  by  case  2. 


Of  Multiplication.  15 

1*.  Simple  quantities  are  multiplied  together  by  uiulti- 
plying  the  coefficients  one  into  the  other^  and  to  the  pro- 
duct aniiexing  the  quantity  which^  according  to  the  ynt- 
thod  of  notation^  expresses  the  product  of  the  species;  pre- 
fixing  the  sign  +  or  — ,  according  as  the  signs  of  the  given 
quantities  are  like  or  unlike* 
Thus         2a        Also  %ah        And  11^^ 

mult,  by     2>h         mult,  by     Sc mult,  by       7ah 

makes        ^ah.      makes        30abc.    makes        77aabdf 


Thus,  by  way  of  illustration,  abcde  will  appear  to  be 
=  abced^  &c.  For,  the  former  of  these  being  equal  to 
every  other  product  of  the  class,  or  termination  e  (by 
hypothesis  and  equal  multiplication),  and  the  latter  equal 
to  every  other  product  of  the  class,  or  termination  d ;  it 
is  evident,  therefore,  that  all  the  products  of  different 
classes,  as  well  as  of  the  same  class,  are  mutually  equal  to 
each  other. 

So  far  relates  to  the  first  general  observation :  it  re- 
mains to  prove  that  abed x pqrst  is  =a  X  b  X  c  X  d X p  X 
q  X  r  X  s  X  t.  In  order  to  which,  let  abed  be  denoted  by 
:> ;,  then  will  abed  x  pqrst  be  denoted  by  :v  X  pqrst ^  or  pqrst 
X  X  (by  case  l),  that  is,  hy  pxqXrXsXtXx-,  which 
1!^  equal  to  x  X  p  X  q  X  r  X  s  X  ty  or  a  xbxcxdxpXq 
X  ^  X  s  X  tj  by  the  preceding  demonstratioiu 

The  reason  of  rule  1  depends  on  tjiese  two  general 
observations :  for  it  is  evident  from  hence,  that  2a  x  ^b 
(in  the  first  example)  is  =2x«XoX^  =  2x 
3X«X^=6xaX^  =  ^ab :  and,  in  the  same 
manner,  11  adf  x  Tab  (in  the  third  example)  appeals 
to  be  =:  11  X  «  X  dxj  X  7  X  a  X  b  =  11  x  7  X  a  x 
axbxdxf^77x  aabdf  =  77  aabdf  But  the 
grounds  of  the  method  of  proceeding  may  be  other- 
wise explained,  thus;  it  ha3  been  observed  that  ab 
(according  to  the  method  of  notation)  defines  the  pro- 
duct of  the  species  a,  b  {in  the  first  example),  therefore 
the  product  of  a  by  3^,  which  must  be  three  times  as 
great  (because  the  multiplier  is  here  three  times  as  great^. 


10  &f  Multiplication. 

In  the  preceding  examples^  all  the  products  are  q^r- 
mative^  the  quantities  given  to  be  muhiplied  being  so ; 
but,  in  those  that  follow,  some  are  affirmative^  and  others 
negative^  according  to  the  different  cases  specified  in  the 
fetter  part  of  the  rule ;  whereof  the  reasons  will  be  ex- 
plained hereafter. 


Mult. 

-f    5a            Mult.     —    5a            Mult. 

—   Sa 

by 

—    e^b            by            4..    p^b            by 

^    6b 

Prod. 

—  ^Oab.         Prod.     -^  ^oab.         Prod. 

+  30abl 

Mult. 

-f-  7'v  ax            Mult.  —  7a'^  au   f   xx 

^y 

—  5\  cy             by        —  6b\  ac  — vy 

Prod. —  35  X  ^'ax  x  Vcz/.  Prod.-f-42aZ>x  V  aa^f  :>f^xV«a-z/y 

In  the  two  last  examples,  and  all  others  where  radi- 
cal quantities  are  concerned,  every  such  quantity  may 
be  considered,  and  treated  in  all  respects  as  a  simple 
quantity,  expressed  by  a  single  letter  ;  since  it  is  not  the 
form  of  the  expres3ion,  but  the  value  of  the  quantity  that 
is  here  regarded. 

2°.  A  fraction  is  multiplied  by  multiplying  the  nume- 
rator thereof  by  the  given  multiplier^  and  making  the  pro- 
duct  a  numerator  to  the  given  denominator. 

rr»        a  ,       ac         .       Sac        ^    ,       ,        6aacd 

ihus  -7-  X  c  makes  -7- ;     also  ---  x  2ad  makes  ; 

b  b  b  b 


will  be  truly  defined  by  3<2^,  or  ab  taken  three  times  :  bu^ 
since  the  product  of  a  by  2>b  appears  to  be  Zab^  it  is  plain 
that  the  product  of  2rt  by  Zb  must  be  twice  as  great  as 
that  of  a  by  3^,  and  therefore  will  be  truly  expressed  by 
6a^.  Thus,  also,  the  product  of  the  species  ah  and  c,  in 
the  second  example,  being  abc  by  bare  notation,  it  is  evi^ 
dent  that  the  product  of  6ab  by  c  will  be  truly  defined  by 
6abc^  or  acb  six  times  taken,  and  consequently  the  product 
of  6ab  and  5c ^  by  SOabc^  or  6abc  taken  five  times,  the  mul- 
tiplier here  being  five  times  as  great. 

The  reason  of  rule  2°  may  be  thus  demonstrated  :  let 
the  numeratgr  of  any  proposed  fraction  be  denoted  by  A» 


Of  Multiplication.  17 

likewise  -^  X  7Vax  makes ^  ;  lastly,  ^ 


X/2ab  makes  

Vaa  +  XX 

3°.  Fractions  are  multiplied  i?ito  07ie  another  by  multi- 
plying the  numerators  together  for  a  nexu  numerator^  and 
the  denominators  together  for  a  nexv  denominator. 

rjy.         a       c       ac     2ab      Sad      10a%d 

21xy      SaVa  __  eSaxyVg      — 5aVx       —  2a  _ 
Vaz  S^  8bV^     '        Sbc ""  b     ^ 

lOaW^ .  and  ^^^^       5bVaa  +  xx  _ 
3bbc     '  v^  a  -{-"z        "^ 

15ab  X  ^xy  X  *>/aa  4.  xx 
a+zx  S/ab 


the  denominator  by  B,  and  the  given  multiplicator  by  C  : 

then,  I  say,  that  -— -  is  equal  to  --  x  C.     For,  since  -— 

denotes  the  quantity  which  arises  by  dividing  AC  by  B, 

and  ~  the  quantity  which  arises  by  dividing  A  by  B,  it  is 

evident  that  the  former  of  these  two  quantities  must  be 
C  times  as  great  as  the  latter,  because  the  dividual  is  C 
times  as  great  iu  the  one  case  as  in  the  other ;  and  there- 
fore must  be  equal  to  the  latter  C  times  taken,  that  is, 

AC  A 

-J-  must  be  equal  to  -^  X  C,  as  was  to  be  shown. 
JtS  B 

The    reason    of    rule    3°    will    appear    evident    from 

the  preceding  demonstration  of   rule  2"^.       For,  it  be- 

A  AC 

ing  there  proved  that  --  x  C  is  equal  to   ----,  it  is  ob- 
t%.  B  B 

A      C  AC 

vious  that  ~  x  r—  can  be  only  the  D  part  of  —  ;   be- 

D 


,18  Of  Multiplication. 

4°.  Surd  quantities  under  the  same  radical  sign  are  mul- 
tiplied like  rational  quantities^  only  the  product  must  stand 

under  the  same  radical  sign.  

Thus,  VF  X  VF  =  V35  \^Va  X  J^b  =  Vab ; 
\/7fc  X  i^5ad  z=l25abcd;  Wab  X  5Vc  =  \S\^abc  ; 
2aV2cy    X     3b  V  5  ax    (z=:    eab    X    V2cy    X     S^Sax) 

^   J    , ,    "Tab      Isx  ^     5c      JlSd 

=  eabVlOacxy ;    and  V —  X  -%  V— r  = 

^  5x   ^  3a        9d^  2b 

35abc     hoAdx 

45dx  ^    6ab 

C 

cause   --,  the  multiplier  here,  is  but  the  D  part  of  the 

AC 

former  multiplier  C :    but  r^^  is  also  equal  to  the  D 

AC 

part  of  the  same  ~— -  ;    because  its  divisor  is  D  times 
B 

AC 

as  great  as  that  of  -— - :    therefore  these  two  quantities, 
13 

A      C         AC 

--  X  -rr  and  -—--,  being  the  same  part  of  one  and  the  sam^ 
r>      D         BD 

quantity,  must  necessarily  be  equal  to  each  other  ;  which 
zvas  to  be  proved.  ^ 

As  to  rule  4°  for  the  multiplication  of  similar  ra- 
dical quantities,  it  may  be  explained  thus :  suppose 
VA  and  VB  to  represent  the  two  given  quantities  to 
be  multiplied  together ;  let  the  former  of  them  be  de- 
noted by  a^  and  the  latter  by  ^,  that  is,  let  the  quan- 
tities represented  by  a  and  b  be  such,  that  aa  may  be  = 
A,  and  ^/^  =  B  ;  then  the  product  of  VA  by  VB^  or 
of  a  by  ^,  will  be  expressed  by  ab^  and  its  square  by 
ab  X  abi  but  ab  x  ab  i^  -=1  a  X  b  X  a  X  b -=  aa  X  bb  (by 
the  general  pbservations  premised  at  the  beginning  of 
this  section)  ;  whence  the  square  of  the  product  is  like- 
wise truly  expressed  by  aa  X  bb^  or  its  equal  A  X  B  ;  and 
consequently  the  product  itself  by  VA  X  B,  that  is,  by 
the  quantity  which,  being  multiplied  into  itself,  produces 
A  X  B. 


Of  Multiplication.  1 9 

4^°.  Powers^  or  roots  ^  of  the  same  quantity  are  77iultiplied 
together  by  adding  their  exponents:  but  the  exponents 
here  understood  are  those  defined  in  p.  5,  where  roots  are 
represented  as  fractional  powers. 

Thus,  x^  X  ^Ms  =  ^*  ;  a  +  zf  X  a  +  z^  =  a  +  z]'  i 

x'^  X  x^  ^  X  =  a:^  ;   and  x^  X  x    =:  x'^  z=:  x  ;   also 

aa  -f  22*]^  X  «a  -f  zz*]^  is  =  aa  4-  zz^  ^  =  aa  +  22  ;  and 


In  the  same  manner,  the  product  of  >/A  X  V'B  will 
appear  to  be  VAB  :  for  if  VA  be  denoted  by  cr,  and 
\/B  by  b ;  or,  which  is  the  same  thing,  if  aaa  =  A,  and 
W^  =  B  ;  then  will  VA  X  v'B^=  a  x  <^  (or  ab)  and 
its  cube  z=z  ab  X  ab  X  ab  =  aaa  X  bbb  =  AB  (by  the 
aforesaid  observations)  ;  whence  the  product  itself  will 
evidently  be  expressed  by  VAB. 

^  The  grounds  of  these  operations  may  be  thus 
explained.  First,  when  the  exponents  are  whole  num- 
bers, as  in  example  1,  the  demonstration  is  obvious, 
from  the  general  observations  premised  at  the  begin- 
ning of  the  section:  for,  by  what  is  there  shown, 
x^  X  x^y  or  XX  X  XXX  is  =  xX^XxX^X^^^^  (by 
notation).  But  in  the  last  example,  where  the  expo- 
nents are  fractions,  let  c  +  z/"j^  be  represented  by  x  j 
that  is,  let  the   quantity  x  be  such,  that  :v  X  ^  X  -v  x 

a:  X  ^  X  ^,  or  ^^  may  be  equal  to  c  +y;  so  shall  c  -f  z/ 1 
be  expressed  by  :i^^ ;    because,  by  what  has  been  already 
shown,  x^  X  x^    is  =    x^ :     and    in    the  same  manner 

will  c  +  2/"]  3  be  expressed  by  x^  ;  because  x^  x  x^  X  ^  is 

likewise  =  x^.       Therefore   c  +  y"]     X   c  -f-  z/"]^   is  = 

x^  X  x^  =  x^  =  the  fifth  power  of  c  -|-  t/]^^    which  is 

c  +  y'\^^  by  notation. 


20  Of  Multiplication. 

6°.  A  compound  quantity  is  multiplied  by  a  simple  07ie^ 
by  multiplying  every  term  of  the  multiplicand  by  the  mul- 
tiplier. 


Thus       a  +  2b"^Zc          Also         a2  —  5aVx'\-7b 
mult,  by         3(2 mult,  by   8c 


makes     Sa^-f-Grt^  —  9ac;  makes    8a^c — 40acV x '\- 5^bc ; 
And       5a^  ^-^8ab  +  6aC'—7bc+12b^  —  9c^  ~ 

mult,  by  Sabc 
makes  15a^bc — 24a^b^C'\'18a^bc^ — 21  ab^c^+36ab^C'''27abc^. 


To  explain  the  reason  of  the  two  last  rules,  let  it  be 
first  proposed  to  multiply  any  compound  quantity,  as 
a  +  b  —  c  —  dy  by  any  simple  quantity  f;  and,  I  say, 
the  product  will  be  ctf  +  bf  —-  cf  —  df  For  the  pro- 
duct of  the  affirmative  terms,  a  +  b^  will  he  of  +  bf 
because  to  multiply  one  quantity  by  another  is  to  take  the 
multiplicand  as  many  times  as  there  are  units  in  the 
multiplier,  and  to  take  the  whole  multiplicand  (a  +  /;) 
any  number  of  times  (f)^  is  the  same  as  to  take  all  its 
parts  (a,  b)  the  same  number  of  times,  and  add  them 
together.  Moreover,  seeing  a  +  b  —  c  ^-^  d  denotes  the 
excess  of  the  affirmative  terms  (a  and  b)  above  the 
negative  ones  (c  and  d)y  therefore,  to  multiply  a  +  b  — 
c  —  d  by  f  is  only  to  take  the  said  excess  f  times  ;  but 
f  times  the  excess  of  any  quantity  above  another  is, 
manifestly,  equal  to  f  times  the  former  quantity,  minus 
f  times  the  latter ;  but  f  times  the  former  is,  here, 
equal  to  qf  +  bf  (by  what  has  been  already  shown), 
and/times  the  latter,  for  the  same  reason,  will  be  equal 
to  cf  +  df  and  therefore  the  product  oi  a  +  b  —  c  —  d 
by  f  is  equal  to  af  +  bf  —  cf  —  df;  as  was  to  be 
proved.  Hence  it  appears,  that  a  compound  quantity  is 
multiplied  by  a  simple  affirmative  quantity,  by  multi- 
plying every  term  of  the  former  by  the  latter,  and  con- 
necting the  terms  thence  arising  with  the  signs  of  the  mul- 
tiplicand. 


Of  Multiplicatmu  21 

7°,  Compound  quaiitities  are  multiplied  into  one  another 
by  multiplying  every  term  of  the  multiplicand  by  each  term 
of  the  multiplier^  successively^  and  connecting-  the  several 
products  thus  arising  with  the  signs  of  the  multiplicand^  if 
the  multiplying  term  be  afirmative^  but  with  contrary  signs ^ 
if  negative. 

Thus  the  product  of  5a    +    Sx 

multiplied  by  Sa    +    2x 

.„  ,  r     13aa  -f    9ax                  1 

^^^^1^^  1              +10ax  +  6xx     S 

which  contracted  by  unit-    f     i^aa  +  19ax  +  6xx. 
ing  the  like  terms,  is         (. 


But  to  prove  that  the  method  also  holds  good  when  both 
the  quantities  are   compound  ones,  let  it  be   now  pro- 
posed to  multiply  A  —  B  by  C  —  D  ;  then,  I  say,  the  pro- 
duct will  be  truly  expressed  by  AC  —  BC  —  AD+  BD, 
For,  it  has  been  already  observed,  that  to  multiply  one 
quantity    by    another    is    to    take    the    multiplicand    as 
many  times  as  there  are  units  in  the  multiplier ;    and, 
therefore,    to    mutiply    A  —  B    by    C  —  D    is  only  to 
take  A  —  B  as  many  times  as  there  are  units  in  C  —  D  : 
Now    (according    to    the    method   of   multiplying  com- 
pound quantities)  I  first  take  A  —  B,  C  times  (or  multiply 
by  C),  and  the  quantity  thence  arising  will  be   AC  — 
BC,  by  what    is    demonstrated  above*       But   I   was  to 
have  taken  A  —  B  only  C  —  D  times  ;  therefore,  by  this 
first  operation,    I    have    taken  it  D  times    too    much ; 
whence,   to   have  the  true  product,   I  ought  to  deduct 
D  times    A  —  B    from    AC  —  BC,  the  quantity  thus 
found  ;   but  D  times  A  —  B,  by  what  is  already  proved^ 
is  equal  to   AD  —  BD  ;     which  subtracted    from   AC 
—  BC,  or  written  down  with  its  signs  changed,  gives 
the  true  product,  AC  —  BC  —  AD-f-BD,   as  was  to 
be  demonstrated.      And,  universally^  if  the  sign  of  any 
proposed    term  of    the    multiplier,    in    any   case    what- 
ever, be  affirmative,  it  is  easy  to  conceive  that  the  re- 
quired product  will  be  greater  than  it  would  be  if  there 


Of  Multiplication*, 

Likewise  the  product 
of         c^  +  a^h  +  alP'  +  ly^ 
by         a^h   


fa^+a^l?  +  a^^+aP  \ 

^s       \     ^a^b  —  aW  —  ab^^b^S 
Which,    by   striking  out  the  terms  that  destroy  one 
another,  becomes  a'*  —  b^* 


were  no  such  term,  by  the  product  of  that  term  into 
the  whole  multiplicand  -,  and  therefore  it  is,  that  this 
product  is  to  be  added,  or  written  down  with  its  proper 
signs,  which  are  proved  above  to  be  those  of  the  multi- 
plicand. But  if,  on  the  contrary,  the  sign  of  the 
term  by  which  you  multiply  be  negative ;  then,  as 
the  required  product  must  be  less  than  it  would  be  if 
there  were  no  such  term,  by  the  product  of  that  term 
into  the  whole  multiplicand,  this  product,  it  is  manifest, 
ought  to  be  subtracted,  or  written  down  with  contrary 
signs. 

Hence  is  derived  the  common  rule,  that  like  sig?is  pro- 
duce +,  and  unlike  signs^  — . 

For,  first,  if  the  signs  of  both  the  quantities,  or  terms, 
to  be  multiplied  be  affirmative,  and  therefore  like^  it  is 
plain  that  the  sign  of  the  product  must  likewise  be  affir- 
mative. 

Secondly,  also,  if  the  signs  of  both  quantities  be  nega- 
tive, and  therefore  still  like^  that  of  the  product  will  be 
affirmative,  because  contrary  to  that  of  the  multiplicand^ 
by  rvhat  has  been  just  now  proved. 

Thirdly,  but  if  the  sign  of  the  multiplicand  be  affirma- 
tive, and  that  of  the  multiplier  negative,  and  therefore 
unlike^  the  sign  of  the  product  will  be  negative,  because 
co7itrary  to  that  of  the  multiplicand. 

Lastly,  if  the  sign  of  the  multiplicand  be  negative,  and 
that  of  the  multiplier  affirmative,  and  therefore  still  unlike^ 
the  sign  of  the  product  will  be  negative,  because  the  same 
zvith  that  of  the  multiplicand. 

And  these  four  are  all  the  cases  that  can  possibly  hap'- 
pen  with  regard  to  the  variation  of  signs. 


Of  Multiplicatio7i. 


2Z 


Other  examples  in  multiplication,  for  the  learner's  ex- 
ercise, may  be  as  follows ;  from  which  he  may,  if  he  please, 
proceed  directly  to  division,  by  passing  over  the  interven- 
ing scholium. 


1. 

Multiply 
by 

product 

Multiply 
by 

product 

Multiply 
by 

product 

Multiply 
by 

product 

x^+xy  +y^ 
x^  —  xy  +y^ 

x^+x'y  +  x^y^ 

—  x^y — x^y^  —  xy^ 

+  x^y^  +  xy^  +  y^ 

2. 

2^2 —    Sax   +    4x^ 
5a^ —    ^ax  —   2x^ 

—    4a'x^  +    6ax^  —  8;c^ 

3. 

lOa'^  —  ^ra^x  +  34aV  —  \Mx^  —  S^v^ 

2a   —    2b    +    2c 
2a    —   Ab    +    5c 

6aa  —   4ab  -f-    4ac 

—  12ab+    Sbb—    Sbc 

+  15ac — lObc  +  lOcc 

4. 

6aa —  Idab  -f-  19ac  -f-     Sbb--18bc  +  lOcc. 

o?  —  Sd^b+    Sab^^b^ 
a^  —  2ab  +    b^ 

aS  —  Sa^b+    3a^^~      aH^ 

—  2a*^+    ^aH^—    6a^'-  +  2ab^ 

+         ^3^2  _      3^2/,3  ^  3^^4  _  1,5 

a^  — .  sa^'b  +  lOa^b^  —  lOd'b^  +  Sab""  —  b\ 
SCHOLIUM. 

The  manner  of  proceeding,  in  referring  the  reasons 
of  the  different  cases  of  the  signs  to  the  multiplication 
of  compound  quantities,  may  perhaps  be  looked  upon 
as  indirect,  and  contrary  to  good  method ;  according  to 
which,  it  may  be  thought  tibat  these  reasons  ought  to 


24  Of  Multiplication^ 

have  been  given  before,  along  with  the  rules  for  simple 
quantities,  as  this  is  the  way  that  almost  all  authors  on  the 
subject  have  followed. 

But,  however  indirect  the  method  here  pursued  may 
seem,  it  appears  to  me  the  most  clear  and  rational ;  and 
I  believe  it  will  be  found  very  difficult,  if  not  impossible, 
without  explaining  the  rules  for  compound  quantities 
first,  to  give  a  learner  a  distinct  idea  how  the  product 
of  two  simple  quantities  with  negative  signs,  such  as 
—  h  and  — c,  ought  to  be  expressed,  when  they  stand 
alone,  independent  of  all  other  quantities :  and  I  can- 
not help  thinking  farther,  that  the  difficulties  about  the 
signs,  so  generally  complained  of  by  beginners,  have 
been  much  more  owing  to  the  manner  of  explaining 
them,  this  way,  than  to  any  real  intricacy  in  the  sub- 
ject itself;  nor  will  this  opinion,  perhaps,  appear  ill 
grounded,  if  it  be  considered  that  both  — a  and  — ^, 
as  they  stand  here  independently,  are  as  much  im- 
possible in  one  sense,  as  the  imaginary  surd  quantities 
V —  b  and  s/ —  c  ;  since  the  sign  — ,  according  to 
the  established  rules  of  notation,  shows  that  the  quan- 
tity to  which  it  is  prefixed  is  to  be  subtracted ;  but,  to 
subtract  something  from  nothing  is  impossible,  and  the 
notion  or  supposition  of  a  quantity  less  than  nothing, 
absurd  and  shocking  to  the  imagination :  and,  cer- 
tainly, if  the  matter  be  viewed  in  this  light,  it  would 
be  very  ridiculous  to  pretend  to  prove,  by  any  show  of 
reasoning,  what  the  product  of  —  h  hy  —  c,  or  of 
v' —  b  by  S/ —  c,  must  be,  when  we  can  have  no 
idea  of  the  value  of  the  quantities  to  be  multiplied. 
If,  indeed,  v/e  were  to  look  upon  —  h  and  —  c  as  real 
quantities,  the  same  as  represented  to  the  mind  by  b 
and  c  (which  cannot  be  done  consistently,  in  pure  alge- 
bra, where  magnitude  only  is  regarded),  we  might  then 
attempt  to  explain  the  matter  in  the  same  manner  that 
some  others  have  done  ;  from  the  consideration,  that, 
as  the  sign  —  is  opposite  in  its  nature  to  the  sign  +, 
it  ought  therefore  to  liave,  in  all  operations,  an  oppo- 
site effect ;   and,  conseqn.ently,  that  as  the  product,  when 


Of  Multiplication.      '  25 

the  sign  +  is  prefixed  to  the  multiplier,  is  to  be  added,  so, 
on  the  contrary,  the  product,  when  the  sign  —  is  prefixed, 
ought  to  be  subtracted. 

But  this  way  of  arguing,  however  reasonable  it  may 
appear,  seems  to  carry  but  very  little  of  science  in  it, 
and  to  fall  gready  short  of  the  evidence  and  conviction 
of  a  demonstration :  nay,  it  even  clashes  widi  first 
principles,  and  the  more  established  rules  of  notation ; 
according  to  which  the  signs  +  and  —  are  relative  only 
to  the  magnitudes  of  quantities,  as  composed  of  diffe- 
rent terms  or  members,  and  not  to  any  future  opera- 
tions to  be  performed  by  them  :  besides,  when  we  are 
told  that  the  product  arising  from  a  negative  multiplier 
is  to  be  subtracted,  we  are  not  told  what  it  is  to  be  sub- 
tracted from  ;  nor  is  there  any  thing  from  whence  it  can 
be  subtracted,  when  negative  quantities  are  independent- 
ly considered.  And,  farther,  to  reason  about  opposite 
effects,  and  recur  to  sensible  objects  and  popular  consi- 
derations, such  as  debtor  and  creditor,  tPc.  in  order  to 
demonstrate  the  principles  of  a  science  whose  object  is 
abstract  number,  appears  to  me  not  well  suited  to  the 
nature  of  science,  and  to  derogate  from  the  dignity  of 
the  subject. 

It  must  be  allowed,  that,  in  the  application  of  alge- 
bra to  different  branches  of  mixed  mathematics,  where 
the  consideration  of  opposite  qualities,  effects,  or  posi- 
tions can  have  place,  the  usual  methods  have  a  better 
foundation ;  and  the  conception  of  a  quantity  abso- 
lutely negative  becomes  less  difficult.  Thus,  for  ex- 
ample, a  line  may  be  conceived  to  be  produced  out, 
both  ways,  from  any  point  assigned ;  and  the  part  on 
the  one  side  of  that  point  being  taken  as  positive^  the 
other  will  be  negative.  But  the  case  is  not  the  same  in 
abstract  number,  whereof  the  beginning  is  fixed  in  the 
nature  of  things,  from  whence  we  can  proceed  only  one 
way. 

There  can,  therefore,  be  no  such  things  as  nega- 
tive numbers,  or  quantities  absolutely  negative,  in  pure 
algebra,  whose  object  is  number,  and  where  every 
muhiplication,    division,    ^c.   is  a  multiplication,    divi- 

E 


26  Of  Multiplication. 

sion,  &fc.  of  numbers,  even  in  the  application  thereof; 
for,  when  we  reason  upon  the  quantities  themselveSj 
and  not  upon  the  numbers  expressing  the  measures  of 
them,  the  process  becomes  purely  geoinetrical^  whatever 
symbols  may  be  used  therein,  from  the  algebraic  notation  ; 
which  can  be  of  no  other  use  here  than  to  abbreviate  the 
work. 

However,  after  all,  it  may  be  necessary  to  show  upon 
what  kind  of  evidence  the  multiplication  of  negative 
and  imaginary  quantities  is  grounded,  as  these  some- 
times occur,  in  the  resolution  of  problems :  in  order 
to  which  it  will  be  requisite  to  observe,  that,  as  all  our 
reasoning  regards  real^  positive  quantities,  so  the  alge- 
braic expressions,  whereby  such  quantities  are  exhi- 
bited, nuist  likewise  be  real  and  positive.  But,  when 
the  problem  is  brought  to  an  equation,  the  ease  may 
indeed  be  otherwise  ;  for,  in  ordering  the  equation,  so 
much  may  be  taken  away  from  both  sides  thereof,  as  to 
leave  the  remaining  quantities  negative  ;  and  then  it  is, 
chiefly,  that  the  multiplication  by  quantities  absolutely 
negative  takes  place. 

Thus,  if  there  were  given  the  equation  a =  r, 

in  order  to  find  x  ;    then,  by  subtracting  the  quantity  a 

from  each  side  thereof,  we    shall  have  —  --   =  c  —  a; 

b 

which  multiplied  by  —  ^,  according  to  the  general  rule^ 

gives  X  :=z  —  cb  +  ab ;    that  is,  —  --  by  —  b  will  give 

+  X  ;  c  by  —  b,  —  cb  ;  and  —  a  by  —  b,  -j- ab  ;  which 
appear  to  be  true  ;  because  the  products  being  thus  ex- 
pressed, the  same  conclusion  is  derived,  as  if  both  sides 
of   the  original  equation    had    been    first    increased  by 

c,  and  then  multiplied  by  b  ;    where  both  the  mul- 

b 

tiplier  and  multiplicand  are  real,  affirmative  quantities, 

and  where  the  whole  operation  is,  therefore,  capable  of 

a  clear  and  strict  demonstration:    but  then  it  is  not  in 

consequence  of  any  reasoning  I  am  capable  of  forming 


Of  Multiplication.  27 

about  —  -  and  •—  b^  or  about  +  c  and  —  b^  considered 
b 

independently,  that  I  can  be  certain  that  their  product 

ought  to  be  expressed  in  that  manner. 

So,  likewise,  if  there  were  given  the  equation  a  — 

—  =  c ;  by  transposing  a,  and  taking  the  square  root 
b  _ 

f  •    ^2  

on  both  sides,  we  shall  have  y — ^  =  Vc  —  a;    and 

this  multiplied  by  V — by  will  give  V  x^  (or  x)  =; 
V —  cb  +  ab:  which  also  appears  to  be  true,  because 
the  result,  this  way,  comes  out  exactly  the  same  as  if 
the  operations  for  finding  x  had  been  performed  alto- 
gether by  real  quantities :  but,  notwithstanding  this, 
it  is  not  from  any  reasoning  that  I  can  form  about  the 

j        ^» 
multiplication    of    the     imaginary    quantities    y f 


and  V —  by  &c.  considered  independently,  that  I  can 
prove  their  product  ought  to  be  so  expressed  ;  for  it  would 
be  very  absurd  to  pretend  to  demonstrate  what  the  pro- 
duct of  two  expressions  must  be,  which  are  impossible 
in  themselves,  and  of  whose  values  we  can  form  no  idea. 
It  indeed  seems  reasonable,  that  the  known  rules  for  the 
signs,  as  they  are  proved  to  hold  good  in  all  cases  what- 
ever, where  it  is  possible  to  form  a  demonstration,  should 
also  answer  here :  but  the  strongest  evidence  we  can  have 
of  the  truth  and  certainty  of  conclusions  derived  by  means 
of  negative  and  imaginary  quantities,  is  the  exact  and  con- 
stant agreement  of  such  conclusions  with  those  determin- 
ed from  more  demonstrable  methods,  wherein  no  such 
quantities  have  place. 

In  the  foregoing  considerations,  the  negative  quali- 
ties —  by  —  c,  &c.  have  been  represented,  in  some 
cases,  as  a  kind  of  imaginary  or  impossible  quantities ; 
it  may  not,  therefore,  be  improper  to  remark  here,  that 
such  imaginary  quantities  serve,  many  times,  as  much 
to  discover  the  impossibility  of  a  problem,  as  imaginar}- 


28  Of  Division* 

surd  quantities  :  for  it  is  plain,  that,  in  all  questions  re- 
lating to  abstract  numbers,  or  such  wherein  magnitude 
only  is  regarded,  and  where  no  consideration  of  posi- 
tion, or  contrary  values,  can  have  place  ;  I  say,  in  all 
such  cases,  it  is  plain  that  the  solution  will  be  altogether 
as  impossible,  when  the  conclusion  comes  out  a  negative 
quantity,  as  if  it  were  actually  affected  with  an  imagi- 
nary surd  ;  since,  in  the  one  case,  it  is  required  that  a 
number  should  be  actually  less  than  nothing ;  and,  in 
the  other,  that  the  double-rectangle  of  two  numbers 
should  be  greater  than  the  sum  of  their  squares  ;  both 
which  are  equally  impossible:  but,  as  an  instance  of 
the  impossibility  of  some  sort  of  questions,  when  the 
conclusion  comes  out  negative,  let  there  be  given,  in 
a  right-angled  triangle,  the  sum  of  the  hypothenuse 
and  perpendicular  =  ^,  and  the  base  =  h^  to  find 
the  perpendicular ;  then,  by  what  shall  hereafter  be 
shown  in  its  proper    place,    the  answer  will  come  out 

,  and  is  possible,  or  impossible,  according  as  the 

quantity  —    is  affirmative  or  negative,   or  as  a  is 

greater  or  less  than  h  ;  which  will  manifestly  appear  from 
a  bare  contemplation  of  the  problem  :  and  the  same  thing 
might  be  instanced  in  a  variety  of  other  examples. 


SECTION  V. 

Of  Division. 


DIVISION  in  species,  as  in  numbers,  is  the  converse 
of  multiplication,  and  is  comprehended  in  the  seven  fol- 
lowing cases. 

1°.  JVhen  one  simple  quantity  is  to  be  divided  by  a?i- 
other^  and  all  the  factors  of  the  divisor  are  also  found  iii 
the  dividend^  let  those  factors  be  all  cast  off  or  expunged^  and 
then  the  remaining  factors  of  the  dividend^  joined  together^ 
will  express  the  quotient  sought.     But  it  is  to  be  observed, 


OfDivisioJU  29 

that,  both  here  and  in  the  succeeding  cases,  the  same  rule 
is  to  be  regarded,  in  relation  to  the  signs,  as  in  multipli- 
cation, 'Diz.  that  like  signs  give  +,  and  unlike^  — .  It  may 
also  be  proper  to  observe,  that,  when  any  quantity  is  to  be 
divided  by  itself,  or  an  equal  quantity,  the  quotient  will 
be  expressed  by  a  unit,  or  1. 

Thus  ^  -4-  «  gives  1  ;   and  2ab  -—-  2ab  gives  1 ; 

moreover,  Sabcd-r-  ac  gives  Sbd ; 

and  16bc  -^  8b  gives  2(? :  for  here  the  dividend,  by 
resolving  its  coefficient  into  two  factors,  becomes  2x8 
X  b  X  c  ;  from  whence  casting  off  8  and  ^,  those  common 
to  the  divisor,  we  have  2  x  c,  or  2c.  In  the  same  manner, 
by  resolving  or  dividing  the  coefficient  of  the  dividend  by 
that  of  the  divisor,  the  quotient  will  be  had  in  other  cases  : 
Thus,  20abc^  divided  by  4c,  gives  Sab  ;  and  —  Slab 
\^xif  X  ^xx  -f  i/if^  divided  by  —  1  TaVxi/j  gives  +  Sb 
\/xx  +  yy» 

2°.  But  if  all  the  factors  of  the  divisor  be  not  found  in 
the  dividend^  cast  off  those  {if  any  such  there  be)  that  are 
common  to  both^  and  write  down  the  reinaining  factors  of 
the  divisor^  joined  together^  as  a  deno7ninator  to  those  of 
the  dividend ;  so  shall  the  fraction  thus  arising  express 
the  quotient  sought.  But  if,  by  proceeding  thus,  all  the 
factors  in  the  dividend  should  happen  to  go  off,  or  va- 
nish, then  a  unit  will  be  the  numerator  of  the  fraction  re- 
quired. 

Thus,  abCy  divided  by  bcd^  gives  -: 

^ax 
And  16a^bx^^  divided  by  Sabcx^^  gives  ^ — : 

The  first  rule,  given  above,  being  exactly  the  con* 
verse  of  rule  1°  m  the  preceding  section,  requires  no 
.other  demonstration  than  is  there  given.  The  second 
>'ule  (as  well  as  those  that  follow  hereafter  upon  frac- 
tions) depends  on  this  principle,  that,  as  many  times 
as  any  one  proposed  quantity  is  contained  in  another, 
just  so  many  times  is  the  half,  third,  fourth,  or  any  other 
assigned  part  of  the  former,  contained  in  the  half,  third, 
fourth,  or  other  corresponding  part  of  the  latter ;    and 


30  Of  Division. 

Likewise,  QTabVxy^  divided  by  9aWxy^  gives  £- : 

a 

And  SabVai/^  divided  by  16a^bVa^^  gives  — .. 

2a 

3°.  One  fraction  is  divided  by  another ^  by  multiplying 
the  denominator  of  the  divisor  into  the  numerator  of  the 
dividend  for  a  new  numerator^  and  the  numerator  of  the 
divisor  into  the  denominator  of  the  dividend  for  a  new  de- 
nominator. 

1  hus,  y,  divided  by  -,  gives  ~~ : 
0  d  be 

* ,       5ax    ,.   .  1    1  ,      ^bc     .         35adx 
Also,— ,  divided  by—,  gives  ^^: 

.     ,  6a^b    ,.   .  ,    ,  ,      5ab^     .        ISa^bx 

And ,  divided  by ,  ogives —-. 

5x'  ^    2x'^         25ab^x 


just  so  niany  times,  likewise,  is  the  double,  triple,  qua- 
druple, or  any  other  assigned  multiple  of  the  former,  con- 
tained in  the  double,  triple,  quadruple,  or  other  corre- 
sponding multiple  of  the  latter.  The  demonstration  of 
this  principle  (though  it  may  be  thought  too  obvious 
to  need  one)  may  be  thus  :  let  A  and  B  represent  any 
two  proposed  quantities,  and  AC  and  BC  their  equimul- 
tiples (or,  let  AC  and  BC  be  the  two  quantities,  and  A 

AC       A 

and  B  their  like  parts)  :     I  say,  then,  that  —77  =  -^  : 

BC        B 

AC 

for  the  multiple  of-——-  by   BC  is  manifestly  =  AC; 
BC 

A  A 

and  -~-  X  BC,  the  multiple  of  ~  by  the  same  BC,  is  = 

o (^!/  ^^^^^  ^  ^^^  tnultiplication^j  =  — rj—  (yid,  p.  14 

and  15)  =  AC  :  therefore,  seeing  the  equimultiples  of  the 
two  proposed  quantities  are  the  same,  the  quantities  them- 
selves must  necessarily  be  equal. 

The  second  rule,  given  above,  is  nothing  more  than 
a  bare  application  of  the  principle  here  demonstrated : 


Of  Division.  31 

But,  in  cases  like  this  last,  where  the  two  numerators, 
or  the  denominators,  have  factors  common  to  both,  the 
conclusion  will  become  more  neat  by  first  casting  oif  such 
common  factors. 

Thus    casting  away  ab  out  of   the    two    numerators, 

and  X  out  of  both  the  denominators,  we  have  —  to  be 

divided  by  — ;    whereof  the  quotient  is  — -  :     in  the 

12ac^         4acx  Sc^         x      .         Q>c^d 

same  manner,  — tt  -^ Tr-,  or  —-.  -4 — .,  gives  — --; 

'    lObb  5bd  2b  d    ^  2bx 

J  QiaUxy        7a^\/xy        Q    ^    7a     .        12b 

and  — -i-— ^  -4 ^V-^9  or  — r-  — -,  crives  — . 

5c  lObc    '       1        2b' ^  7a 

When  either  the  divisor  or  the  dividend  is  a  whole 
quantity^  instead  of  a  fraction,  it  may  be  reduced  to  the 
form  of  a  fraction,  by  writing  a  unit  or  1  under  it. 


since,  by  casting  off  the  factors  common  to  the  dividend 
and  divisor  (as  directed  in  the  rule),  it  is  plain  that  we  take 
like  parts  of  those  quantities  :  therefore  the  quotient  arising 
by  dividing  the  one  part  by  the  other  will  be  the  same  as 
that  arising  by  dividing  one  whole  by  the  other. 

As  to  rule  3°,  wherein  it  is  asserted  that  -—  -^  — -  =  rrjrr^ 

it  is  evident  that  AD  and  BC  are  equimultiples  of  the 

AC  A  . 

given  quantities  --  and  -- ;  because  --  X  BD  is  (by  ride 

2"*  in  rmdtiplication)  =  — jt—  =  AD,  and  --  X  BD  = 

crd 

=  CB  :     whence  it  follows  that  the  quotient  of 

A  C 

-.  divided  by  --  will  be  the  same  with  that  of  AD  di- 
B  D 

AD 

videdL  by  BC  ;    which,  by  notation,  is  -^tti  ^s  was  to 

BC 

be  shown.      The  grounds  of  the  note  subjoined  to  this 

rule  are  these:     by    casting   away  all   factors  common 


32  Of  Division, 

Thus,  — -,  divided  by  7^  (or  ~- ),  giv 


ves  ■ 


35cy/ 

^     ,       o;  ,      5a2|^\    J.   .,    ,1      O^t-^      .        ISa^bif 
And  5a^£>  (or ),  divided  by  -^^^  gives r-^. 

1       /  o2y  \yOCOC 

4°.  iS'z/r^  quantities^  under  the  same  radical  sign^  are 
divided  by  one  another  like  rational  quantities^  only  the 
quotient  must  stand  under  the  given  radical  sig?L 

Thus,  the  quotient  of  Vab  by  Vb    is  Va    : 

That  of  s/l6xxy  by  VSxy  is  V2x  : 

T-u  .    f     hOabb ,         I'sab  .       llOabbc        '    l2b 

1  hat  ot  v/ by  \/  —  is  v r—>  oi^  V — • 

^     Sc       ^  ^    c        ^  \5abc'        >  3 

And  that  of  6abVlOacxy  by  2aV2cy  is  SbVjax. 

5°>  Different poxvers^  or  roots^  of  the  same  quantity  are 
divided  one  by  another^  by  subtracting  the  exponent  of  the 
divisor  from  that  of  the  dividend^  and  placing  the  remainder 
as  an  exponent  to  the  quantity  given.  But  it  must  be  ob- 
served, that  the  exponents  here  understood  are  those  de- 
fined in  p.  5  ;  where  all  roots  are  represented  as  fractional 
powers.  It  will  likewise  be  proper  to  remark  further, 
that,  when  the  exponent  of  the  divisor  is  greater  than  that 
of  the  dividend,  the  quotient  will  have  a  negative  exponent, 
or,  which  comes  to  the  same  thing,  the  result  will  be  a 
fraction,  whereof  the  numerator  is  a  unit,  and  the  deno- 
minator the  same  quantity  with  its  exponent  changed  to 
an  affirmative  one. 

Thus,  x%  divided  by  a:^,  gives  x^ : 

And  «  +  zl^,  divided  by  a  -f  z"]^,  gives  a  -f-  zY  '• 

Likewise,  x^^  divided  by  ;c*,  gives  x^  : 


to  the  two  numerators,  we  take  equal  parts  of  the  quan- 
tities ;  and,  by  throwing  off  the  factors  common  to  both 
denominators,  we  take  equimultiples  of  those  parts. 

The  two  preceding  rules,  being  nothing  more  than 
the  converse  of  the  4th  and  5th  rules  in  multiplication, 
are  demonstrated  in  them:  though,  perhaps,  the  case 
in  rule  5,  where  the  exponent  comes  out  negative,  may 
stand  in  need  of  a  more  particular  explanation.     Accord^ 


Of  Divisioiu  33. 

Moreover,  c  +  yl  ,  divided  by  c  +  i/1  ,  gives  c  +  2/ r  * 

Lastly,  x^^  divided  by  ^*,  gives  x       ,  or  --. 

x^ 

6°.  A  compound  quantity  is  divided  by  a  simple  one  ku 
dividing  every  term  thereof  by  the  given  divisor. 

Thus,  3a^)  Zabc  +  l^abx  —  9aab  (c  +  4;^  —  3a : 

Also,— 5ac)l5a^fc—  12acy^+5ad^  {—3ab  4— i^--,— : 

5  c, 

and  so  of  others. 

7°.  But  if  the  divisor  J  as  well  as  the  dividend^  be  a  com- 
pound quantity  y  let  the  terms  of  both  quantities  be  disposed  in 
order  y  according  to  the  dime?isions  of  some  letter  in  them^  as 
shall  be  judged  most  expedient  ^  so  that  those  terms  may  stand 
first  wherein  the  highest  pozver  of  that  letter  is  involved^  and 
those  next  where  the  next  highest  power  is  involved^  and  so 
on:  this  being  done^  seek  how  many  times  the  first  term  of 
the  divisor  is  contained  in  the  first  term  of  the  dividend^ 
xvhichy  when  founds  place  in  tA^  quotient^  as  i2i  division  in- 
vulgar  arithmetic^  and  then  multiply  the  xvhole  divisor 
thereby^  subtracting  the  product  from  the  respective  terms  of 
the  dividend;  to  the  remainder  bring  doxvn^xvith  their  pro- 
per signs  J  as  many  of  the  nextfolloxving  terms  of  the  divi- 
dend as  are  requisite  for  the  next  operation;  seeking  again 
how  often  the  first  terin  of  the  divisor  is  contained  in  the 
first  term  of  the  remainder^  which  also  write  down  in  your 
quotient^  and  proceed  as  before^  repeating  the  operation  till 
all  the  terms  of  the  dividend  are  exhausted^  and  you  have 
nothing  remaining. 


ing  to  the  said  rule,  the  quotient  of  x^  divided  by  x^  was 

— 2  1 

asserted  to  be  a:       ,  or  — ,     Now,  that  this  is  the  true 
x^ 

value  is  evident;    because  1  and  x^  being  like  parts  of 

x^  and  x^  (which  arise  by  dividing  by  a:^),  their  quotient 

will  consequently  be  the  same  with  that  pf  the  quantities 

themselves. 

F 


34  Of  Division. 

Thus,  if  it  were  required  to  divide  a^  -f  5a^x  +  Sax^^ 
+  x^  by  a  4-  X  (where  the  several  terms  are  disposed  ac- 
cording to  the  dimensions  of  the  letter  d)jl  first  write 
down  the  divisor  and  dividend,  in  the  manner  below, 
with  a  crooked  line  between  them,  as  in  the  division 
of  whole  numbers  ;  then,  I  say,  how  often  is  a  con- 
tained in  <2^,  or  what  is  the  quotient  of  a^  by  a;  the 
answer  is  «^,  which  I  write  down  in  the  quotient,  and 
multiply  the  whole  divisor,  a  +  at,  thereby,  and  there 
arises  a?  -{^a^x  ;  which  subtracted  from  the  two  first  terms 
of  the  dividend,  It^ves  4a^x ;  to  this  remainder  I  bring 
down  -f  5ax^y  the  next  term  of  the  dividend,  and  then 
seek  again  how  many  times  a  is  contained  in  4a^x ;  the 
answer  is  4aXy  which  I  also  put  down  in  the  quotient, 
and  by  it  multiply  the  whole  divisor,  and  there  arises 
4a^x  +  4ax^y  which  subtracted  from  4a^x  +  5ax^  leaves 
ax^j  to  which  I  bring  down  x^^  the  last  term  of  the 
dividend,  and  seek  how  many  times  a  is  contained  in  ax^y 
which  I  find  to  be  x^  ;  this  I  therefore  also  write  down 
in  the  quotient,  and  by  it  multiply  the  whole  divisor  ;  and 
then,  having  subtracted  the  product  from  ax^  +x^y  find 
there  is  nothing  remains  ;  whence  I  conclude,  that  the  re- 
quired quotient  is  truly  expressed  by  ^^  +  4a^  +  x"^.  See 
the  operation. 


a  +  x)  a^  +  Sa^x  -f-  5ax^  +  x^  {or  -j-  4ax  +  ,%^ 

4a^x  4-  5ax^ 
4a^x  •j-4ax^ 


ax^  +  x^ 
ax^  -f-  .y' 


In  the  same  manner,  if  it  be  proposed  to  divide  a^  — 
^a^x-^  lOa^x^  —  lOa^x'^  +  5ax*~x^  by  a^  —2ax+x^y 
the  quotient  will  come  out  a^  -^^  3a^x  -f  3ax^  — >  x\  as 


the  quotient  will  come  out 
will  appear  from  the  process! 


O/Dhismi.  35 

^Sa^x  +  Sax^^x^ 
a^  —  2ax + x^)  a^ — 5a^x  +  lOa^x-  —  lO^^^^  ^  5^^.4  _  ;^5  (^^^s 

—  Za^x-^-   %a^x^ —   3a- A?^ 

'"°"         +   Q^a^x^—'   Ta^x^  +  Sax"^ 

+^S^x^_~_6fx^  +^'^1 


O  O 

So,  likewise,  if  a*  ►—  x^  be  divided  by  «  — -  :>:,  the  quo- 
tient will  he  a^+a^x +  a^x^ +ax^ +x^  ;  as  by  the  work 
will  appear. 

a  _  ^t")  fl*  —  x^  (a^  -{^a^x  -}-  «V  +  o^f^  +  .y* 


a*^- 
c**-- 

—  a^« 

ax*- 
ax"- 

-x' 

-x' 

Moreover,  if  it  were  required  to  divide  a^  — •  Sci^x^  4 
Sa^x^  —  x^  by  a 3  —  ^a^x  +  3ax^  —  x^^  the  process  wil! 
stand  thus : 

^3  _  Sa^x  +\a^--^3a^::^  +  3a^x^ — x^(a^+Sa^x+Sax^+rc' 

Sax-  —  X  3  )a^ — 3a^x  +Za'^x^ — a^x^ 

4. 2,a^x  —  6«^;c2  4-    a^x^  +  Za^x"^ 

^:^a5x  —  9a^x^  +9a^x^—3a^x'^ ^ 

+  3a'^x'^ — 8a^;f3+6a^4 — x^ 

-j-a^AT^ — 3^/^A"*  +  3«;c^ — x^ 

4-  «3 -^3 — Sa^x"^  +3ax^-^x^ 

O  0  0  0 


36  Of  Involution. 

But  it  is  to  be  observed,  that  it  is  not  always  that  the 
work  will  terminate  without  leaving  a  remainder;  and 
then  this  method  is  of  little  use ;  and  in  all  these  cases 
it  will  be  most  commodious  to  express  the  quotient  in 
the  manner  of  a  fraction,  by  writing  the  divisor  under  the 
dividend,  with  a  line  between  them,  as  has  been  shown  in 
the  method  of  notation. 

It  would  be  needless  to  offer  any  thing  by  way  of  de- 
monstration to  the  two  last  rules,  the  grounds  thereof 
being  already  sufficiently  clear  from  what  has  been  de- 
livered in  the  last  section,  and  the  rules  themselves  no- 
thing more  than  the  converse  of  those  there  demonstrated. 
I  shall  here  show  the  reason  why,  in  division,  as  well 
as  in  multiplication,  like  signs  produce  +,  and  unlike  — . 
in  order  thereto,  it  must  first  be  observed,  that,  according 
to  the  nature  of  division,  every  quotient  whatever  mul- 
tiplied by  the  given  divisor,  ought  to  produce  the  given 
dividend  ;  whence  it  is  evident, 

1.  That  +u)  +ab  {+b ;  because  -f  a^  mult,  by  +  ^, 

gives  +  ab  ; 

2.  That  -f  <?)  —  ab  ( —  b  ;  because  -f  (7,  mult,  by  —  ^, 

gives  —  ab ; 

3.  That  —  a)  -f-  ab  ( —  b  ;  because  —  a,  mult,  by  —  b^ 

gives  +  ab ; 

4.  That  —  a)  —  ab  ( —  b  ;  because  —  a,  mult,  by  —  by 

gives  —  ab ; 
And  these  four  are  all  the  cases  that  can  possibly  hap- 
pen in  respect  to  the  variation  of  the  signs. 


SECTION  VL 

Of  Involution. 


INVOLUTION  is  the  raising  of  powers  from  am 
proposed  root,  and  may  be  performed  by  the  following 
rules. 

1°.  If  the  quantity  or  root  proposed  to  be  involved  have 
no  tndexy  that  is^  if  it  be  not  itself  a  poxver  or  surd^  the 


'  Of  Invohuion.  3.7 

fiawer  thereof  will  be  represented  by  the  same  quantity  un- 
der the  given  index  or  exponent. 

Thus,  the  fifth  power  of  a  is  expressed  by  a* ;  and  the 
seventh  power  o£  a  +  zhy  a -{-zy  • 

2°.  But  if  the  quantity  proposed  be  itself  apower^  or  siird^ 
it  will  be  involved  by  multiplying  its  exponent  by  the  expo- 
ntJit  of  the  proposed  power. 

Thus,  the  cube  or  third  power  of  d^  is  a^ ;  the  fifth 
power  of  xMs  ;ci^  ;  the  fourth  power  of  ax  -^-yy'Y  is 
ax  +  yy']  ^^  ;  and  the  third  power  o£  a  —  xlK  is  a  —  x^i* 

3°.  A  quantity  composed  of  several  factors  multiplied 
together^  is  involved  by  raishig  each  factor  to  the  poxver 
proposed. 

Thus,  the  square  or  second  power  of  ab  is  a^b^  i 
the  cube  or  third  power  of  2ab  is  2^a^b^^  or  ^d^b^ ; 
the  fifth  power  of  3  X  aa  —  xx  X  a  +  b  +  c  is 
243  X  aa  —  xx^Y  X  a  +  b  +  c^  ;  and  the  square  ox 
second  power  of  the  radical  quantity  a^  X  a  +  x\^  is 

—-,2 

a  X  a  +  z  1^. 


The  first  of  the  rules  here  given,  being  mere  nota« 
tion,  does  not  require,  nor  indeed  admit  of  a  demon- 
stration. The  second  may  be  explained  thus :  let  A''* 
be  proposed  to  be  raised  to  the  power  whose  exponent 
is  n :  then  I  say,  that  the  power  itself  will  be  truly  ex- 
pressed by  A^" :  for  since,  by  notation,  A^*  is  the  same 
thing  as  A  X  A  x  A  x  A,  ^c,  continued  to  m  factors^ 
this  raised  to  the  nth  power,  or  multiplied  ?i  times, 
will,  by  the  general  observations  at  p.  13,  be  equal  to 
AxAxAxAxAxA,  &?c,  continued  to  7i  times  nt 
factors,  that  is,  to  7nn  factors;  which,  by  notation, 
is  A*"".  3ut  the  same  thing  may  be  otherwise  demon- 
strated, m  a  more  general  manner,  by  means  of  rule  A 


38  Of  Involution* 

4°.  A  fraction  is  involved  hy  raising  both  the  numerator 
and  the  d^iominator  to  the  power  proposed. 

Thus,  the  second  power  of  ~  is  -- ;  the  third  power 

b        ho 

ot  --— .  IS  - — r- ;     the  tourth  power  of is  ; 

\/x           xh        X  x^        x^ 

the  square  of ^,  or  —  is  — ;  the  cube  of  —  is  — ; 

5             5         25  I        a^ 
a^ 

,  ^v      .  ,1  r  (2a  4-  xxl^ .    aa  +  xxY 

and  the  sixth  power  01 L  is -l L. 


When  any  quantity  to  be  involved  has  the  sign  — 
prefixed,  the  power  itself,  if  the  index  be  an  odd  number, 
must  be  expressed  with  the  same  negative  sign ;  but,  if  an 
even  number,  with  the  contrary  sign,  or  +. 

Thus,  the  second  power  of  —  «,  or  —  a  x  —  a,  is 
+  «^,  because  —  into  —  produces  + :     also,  the  cube 


in  multiplication :  for,  since  powers  raised  from  the 
same  root  are  multiplied  by  addition  of  their  indices, 
it  is  evident  that  the  square  of  A'"  (or  A'"  X  A'")  whe- 
ther the  exponent  ra  be  a  whole  number  or  a  fraction, 
will  be  truly  defined  by  A^"* :  whence  it  likewise  ap- 
pears that  the  cube  of  A^  (or  A^"*  x  A^)  will  be  de- 
fined by  A^»* ;  and  the  fourth  power  of  A^  (or  A^'"  x  A"*) 
by  A'*"^,  is^c 

The  reason  of  the  third  rule  is  also  grounded  on 
the  same  general  observatio7is :  for,  in  the  first  ex- 
ample, where  the  square  of  ab  is  asserted  to  be  aH^^ 
we  know  that  square  to  be  ab  X  ab^  by  the  definition  of  a 
square,  which  quantity  is  there  proved  to  be  the  same 
with  axb  X  ax  b^  or  aax  bb.  So,  likewise,  in  the  se- 
cond example,  the  cube  of  2ab^  or  2ab  x  ^ab  x  ^cib^ 
will  bc=:2x<2X/^X2x«X^X2X«X^  =  2x2X 
\lxcixaxaxh^bxbz=:  8Xa^X^^  =  Sa^^.     And 


Of  Irmolutim.  39 

(jf  ,—  CJJ,  or  +  a^  X  —  ci  19.  —  «^,  because  +  into  — 
produces  ~  j  so,  likewise,  the  fourth  po^-er  of  —  a^  or 
—  a^  X  — «  is  +<2*,  and  the  fifth  power,  or  -|-  <!*  x  —  «, 
is  —  a^  &c.  &c.  Hence  it  appears,  that  all  ev^n  powers, 
whether  raised  from  positive  or  negative  roots,,  will  be 
positive. 

5°.  ^antities  compounded  of  several  terms  are  invohpd 
hy  an  actual  multiplication  of  all  their  parts. 

Thus,  \i  a  •\-  h  were  proposed  to  be  invcflved  to  the 
sixth  power;  by  multiplying  a  +  b  into  itself,  we  shall 
first  have  a^  +  2ab  +  b^j  which  is  the  second  power  of 
a  +  b ;  and  this,  again,  multiplied  by  a  +  b^  gives  a^  + 
Cyd^b  +  Zab^  +  ^^,  for  the  third  power  oi  a  +  b:  whence, 
by  proceeding  on,  in  this  manner,  the  sixth  power  of 
a  +  b  will  be  found  to  come  out  a^  +  6a^b  +  15^ ah^ 
+  20aH^  +  ISd^b"^  +  6ab^  +  b^.     See  the  operation. 

a  +  bj  the  root  or  first  power. 

a  -j-  b 
aa  +    ab 
-{-    ab  +  b^ 


a^  -f  2ab  +  b^,  the  square,  or  second  power. 
a    +      b 


tjie  case  will  be  the  same  when  radical  quantities  are 
concerned,    as  in  the  fourth  example :     for  the  square 

of  a^  X  a  +  x]^j  or  a^  X  a  +  x^^  X  a^  x  a  +  .^l^,  is  = 

a^  X  a^  X  a  +  x]^  X  a  +  .r] 3   -,  ^i  ^  ^^  ;>^  'a~+~x]i  X 

a  +  ^l^^ :    but  a^  x  cT   (by  rule   5°  in  multiplication) 

is  =  a^  =  a,  and  a  +  x~\'^  X  a  +  x^'^  =  a  +  x^^  ;  there- 
lore  our  square,  or  its  equal  product,  is  likewise  expressed 

,2 

by  a  X  a  +  x\^. 

The  4th  rule,  or  case,  for  the  involution  of  fractions  is 
grounded  on  rule  3°  in  multiplication,  and  requires  nc> 
other  4emonstration  than  is  there  given. 


40  Of  Involution* 

a^  4-  ^a^b  +   ^ab^  +      b^  the  cube,  or  third  power. 

a  -^b 

eC^  ^AfO^b  +    6a^b^  -f    4<ab^  +  />Vthe  fourth  power," 

^  +^ 

c*  +  4a^^  +    6a'/^^  +    4«2Z>3  +      ab"" 

4-    Q-^^  -f-    4^3/^^  -f    6fl^^^  +    4g^^   +  b' 
a^  +  5a^^  +  XOcv'b'^  +  lOa^^^  +    ^ab"^   +  ^S  the  5th  power. 

a  +b 

A«  +  5a^b  +  10a'' b^  +  lOa^^  +    SaH"  -f    «(^* 

4-    a^b  -f-    5a^^^  -f  lOa^b^  +  10a^Z>^  +  5^^^  +  b^ 
a^  4-  6a*^  4-  ISa^b^  +  20a^^  +  15aH^  +  6ab^  +  ^%  the  6th 
or  required  power  of  a  4-  ^. 

So,  likewise,  if  it  be  required  to  involve  or  raise  a  — ^ 
to  the  sixfh  power,  tjie  process  will  stand  thus  : 
c  —b 


c?  —  2a^  4-      ^^,  second  pow  ear. 
a  ^^b 


a^  _  20.^  +      ab' 

—   a^b+  2ab^  -^      b^ 

a^^3a^b^+  Sab^  —      W,  third  power. 
a  —  b 

^4_4^3<^^  6«V;2_   4a^3   4-      /^'^  fourth  power. 


n 


—  b 


as  —  4>a^b+    6a^^—   4aV  4.    ab^ 

fl*  _  Sa^b  4- 10<^^^^  —  \OaH^  +  Sab''  —  ^%  fifth  power.  ^ 
or  — b 


Of  Involution.  41 

—    a^b+      a''b^  —  \0a^b^-\'\0aH^  —  5a¥'hh^ 
a^  _  e>a^b  +  ISa-^b^  —  ^Oa^b^  +  ISaF-b"^  —  (5ab^  +  b^^  the 
sixth  power  oi  a-^b  ;  and  so  of  any  other. 

But  there  is  a  rule,  or  theorem,  given  by  Sir  Isaac 
Newton^  demonstrated  hereafter,  whereby  any  power 
of  a  binomial  a  +  b^  or  a  —  ^,  may  be  expressed  in 
simple  terms,  without  the  trouble  of  those  tedious  mul- 
tiplications required  in  the  preceding  operations  ;  which 
is  thus : 

Let  n  denote  any  number  at  pleasure ;    then  the  72th 

power    of    a  +  3    will    be    a^  +  ncT^  b  +  — — HI — . 

«-3,„        72.72— -1.72 2        n-3,»    .    72.72 1.72 2.72 3 

a     b^  ■\ ; .  a     b^  -^ 


1.2.3  1,2.  3.  4 

«-*;a      .      72. 72  — -1.72 2.  72'— ^3.  72 4  n-5  ,  .        ^ 

'^     ''  +  1.2.3.4.5 '^     *''    "^^ 

And  the  72th  power  of  «  —  b  will  be  expressed  in  the  very 
same  manner,  only  the  signs  of  the  second,  fourth,  sixth j> 
&?c.  terms,  where  the  odd  powers  of  b  are  involved,  must 
be  negative. 

An  example  or  two  will  show  the  use  of  this  general 
theorem. 

First,  then,  let  it  be  required  to  raise  <2  +  ^  to  the  third 
power.  Here  72,  the  index  of  the  proposed  power,  be- 
ing 3,  the  first  term,  a^^  of  the  general  expression,  is 
equal    to    c? ;     the    second  72a     b  =  Zc^b ;     the    third 

^lliLZZi,  a^-^B^  =  3a^^;     the  fourth  n.n-l.n-<l^ 
1-2  1.2.3 

a^^'b^^b-,     and    the    fifth  ^'^^-^•^-^•^■^^ 


1.2.3    .4 

^  ~  ^^,  &c.  =  nothing*.  Therefore  the  third  power  of  c  +  5 
is  truly  expressed  by  a^  +  Z{^h  +  2>ab^  +  b'^. 

Again,  let  it  be  required  to  raise  a  +  ^  to  the  sixth 
power;  in  which  case  the  index,  72,  being  6,  we  shall, 
by   proceeding  as  in  the   last  example,    have  a**  =r  a^, 

'^  Fot  one  of  the  factors,  ini  ==0.    Ep 

4 

G 


42  Of  Evoluttoji. 

na-^b  =  em%   IlILilL,    /-^^2  ^  15^4^2     g.^.    and 
1.2  ' 

consequently,  «  +  bY  =  «^^  +  6a*^  +  15a^^a  +  20a^<^^ 
+  ISaH"^  +  6a3*  +  b^  ;  being  the  very  sanie  as  was  above 
determined  by  continual  multiplication. 

Lastly,  let  it  be  proposed  to  involve  cc  +  xy  to  the 
fourth  power. 

Here  a  must  stand  for  cc^  b  for  xy^  and  n  for  4  ;  then, 
by  substituting  these  values,  instead  of  a^  b^  and  7Z,  the 
general  expression  will  become  c^  +  Ac^xy  +  Qcf^x^y^^ 
+  4c^x^y^  +  ^^y^y  the  true  value  sought. 

From  the  preceding  operations  it  may  be  observed, 
that  the  uncise,  or  coefficients,  increase  till  the  indices  of 
the  two  letters  a  and  b  become  equal,  or  change  values, 
and  then  return,  or  decrease  again,  according  to  the  same 
order :  therefore  we  need  only  find  the  coefficients  of  the 
first  half  of  the  terms  in  this  manner,  since  from  these  the 
fest  are  given* 


SECTION  VIL 


Of  Evolution* 

EVOLUTION^  or  the  extraction  ofroots^  being  directly 
the  contrary  of  involution^  or  raising  of  powers y  is  per-* 
formed  by  converse  operations^  viz.  by  the  division  of  indi- 
ces^ as  involution  rvas  by  their  multiplication. 

Thus  the  square  root  of  ^^,  by  dividing  the  exponent 
by,  2,  is  found  to  be  x^ ;  and  the  cube  root  of  x^^  by  di- 
viding the  exponent  by  3,  appears  to  be  x^:  moreover, 
the  biquadratic  root  of  a  -f  at")  ^  will  be  a  +  x^  l    and 

the  cube  root  of  aa  +  xxy  will  ht  aa  +  xx\^*  "^. 

In  the  same  manner,  if  the  quantity  given  be  a  fraction  ♦ 
or  consist  of  several  factors  multiplied  together,  its  root 
will  be  extracted,  by  extracting  the  root  of  each  particular 
factor. 


Of  Evolution,  43 

Thus   the    square  root  of   d^b^  will  be  abi     that  ot 

will  be  — :  and  that  of --.-1,.^_L.  will  be 

— ~    2 

■     . :  moreover,  the  square  root  of  aa  —  xx  |- 

4x  CL — 'X 

will  be  aa  — '^:v V  ;     its  cube  root  aa  —  xx\^' ;     apd 

its  biquadratic  root  aa  —  xoc\^  ;  and  so  of  others.  AIL 
which  being  nothing  more  than  the  converse  of  the  ope- 
rations in  the  preceding  section,  require  no  other  demon- 
stration than  what  is  there  given. 

Evolution  of  compound  quantities  is  performed  by  the 
following  rule. 

Firsts  place  the  several  terms ^  whereof  the  given  quan^ 
tity  is  composed^  in  order^  according  to  the  dimensions  of 
some  letter  therein^  as  shall  be  jitdged  most  commodious  ;  then 
let  the  root  of  the  first  term  be  founds  and  placed  in  the 
quotient ;  which  term  being  subtracted^  let  the  first  term  of 
the  remainder  be  brought  down^  and  divided  by  twice  the, 
first  term  of  the  quotient^  or  by  three  times  its  square^  or 
four  times  its  cube^  &c.  according  as  the  root  to  be  ex- 
tracted is  a  square^  cubicj  cr  biquadratic  one^  &c.  and  let 
the  quantity  thence  arising  be  also  rvritten  doxvn  in  the  quo- 
tient^ and  the  whole  be  raised  to  the  second^  third^  or  fourth^ 
&c.  power ^  according  to  the  aforesaid  cases^  respectively y 
and  subtracted  from  the  given  quantity  ;  and^  if^^y  thing 
7'emai?i^  let  the  operation  be  repeated^  by  always  dividing 
the  first  term  of  the  remainder  by  the  same  divisor^  found 
as  above. 

Suppose,  for  example,  it  were  required  to  extract  the 
square  root  of  the  compound  quantity  2ax  +  a^  +  x^ : 
then,  having  ranged  the  terms  in  order,  according  to 
the  dimensions  of  the  letter  «,  the  given  quantity  will 
stand  thus,  a^  +  2ax  +  x^y  and  the  root  of  its  first 
term  will  be  « ;  by  the  double  of  which  I  divide  2aXy 
the  first  of  the  remaining  terms,  and  add  +  x^  the 
quantity  thence  arising  to  a^  already  found,  and  so  have 
a  +  xm  the  quotient ;  which  being  raised  to  the  second 
power,  and  subtracted  from  the  given  quantity,  nothing 


44  Of  Evolution. 

remains ;    tHei!?efoi'e  a  +  at  is  the  square  root  required* 
Sec  the  operation. 

a^-^^ax-^-x^  (a  +  x 
2a)  2ax 

c^  -*>  "lax  +  x^^  second  power  of  a  +  at. 

o~o       oT 

In  like  manner,  if  the  quantity  a^  —  2a^x  +  3a^;c^ 
—  2ax^  +  x^  be  proposed,  to  extract  the  square  root 
thereof ;  the  answer  will  come  out  a^  —  (2^:  +  x^^  as  ap- 
pears by  the  process. 

a^  —  "2.0^ X  +  oc^x^  —  2ax^  +  x^  {a^  —  ax  +  a;^ 
2a^)—2a^x 
a"^  —  2a^x  -f    a^Y^,  second  power  of  a^  —  ax. 

2d^)       2a^x'^^  first  term  of  the  remainder. 
g^  —  2a^x  4-  Sa^x^ — 2ax^  4-  x"^^  square  of  rt^  •—  a:c  +  a:^ 
rr     ~  -        -  0         0 

Agahi,  let  it  be  required  to  extract  the  cube  root  of 
a^  —  Q>a'^x  +  12ax^  —  8^^,  and  the  w^ork  will  stand  thus  : 
a^    — ea-x  +  12ax^  ~Sx^  (^a  —  2x 
3a^)  —  6a^x 

a^    — 6a^x  '{-\2ax'^  —  8^^,  cube  of  a  —  2x. 
'o  O  0  0~ 

Lastly,  let  it  be  required  to  extract  the  biquadratic  root 
of  16^^"^ — 96x^2/ +  216::c2z/3_216;cz/« +8lz/%  and  the 
process  will  stand  as  follows  : 

16x^    —  9Q)X^y  +  216.x'22^^  —  216xy^  +  SU/  (2x  —  Sy 

-16x^    —  96;c3z/-f  216.%'^  —  2iexy^  -f  SJy^ 
O  0  0  0  0 

And,  in  the  same  manner,  the  root  may  be  deter 
mined  in  any  other  case,  where  it  is  possible  to  be  ex- 
tracted ;  but,  if  that  cannot  be  done,  or  if,  after  all,  there 
be  a  remainder,  then  the  root  is  to  be  expressed  in  the 
maimer  of  a  surd,  according  to  what  has  been  already 
shown.  As  to  the  truth  of  the  preceding  rule,  it  is 
too  obvious  to  need  a  formal  demonstration,  every  ope- 


Of  Evolutipn.  45 

ration  being  a  proof  of  itself.  I  shall  only  add  here, 
that  there  are  other  rules  beside^  this,  for  extracting 
the  roots  of  compound  quantities,  which  sometimes 
bring  out  the  conclusions  rather  more  expeditiously;  but 
as  these  are  confined  to  particular  cases,  and  would  take 
up  a  great  deal  of  room  to  explain  in  a  manner  suf- 
ficiently clear  and  intelligible,  it  seemed  more  eligible  to 
lay  down  the  whole  in  one  easy  general  method,  than  to 
discourage  and  retard  the  learner  by  a  multiplicity  of 
rules.  However,  as  the  extraction  of  the  square  root  is 
much  more  necessary  and  useful  than  the  rest,  I  shall 
here  put  down  one  single  example  thereof,  wrought  ac- 
cording to  the  comnlon  method  of  extracting  the  square 
root,  in  numbers  :  which  I  suppose  the  reader  to  be  ac- 
quainted with,  and  which  he  will  find  more  expeditious 
than  the  general  rule  explained  above. 

Examp.  a^  ^  4^3^^.  ^  ^^^%  ^  4^^3  ^  ^a  ^^^2  ^  2<7.v  -f  jc^ 

2a^  +  2ax)    +A:a^x +  ^a^x^  ' 

+  A^a^x  -f  4*a^x^ 
2a^  +4ax+x^)  +2a^x^  +4ax^  +  x-^ 

'^2a^x^  '}-4ax^  +x'^ 
O  0  0 


SECTION  VIII. 

Of  the  Reduction  of  Fractional  and  Radical  Quan- 
tities. 

THE  reduction  of  fractional  and  radical  quantities  is 
of  use  in  changing  an  expression  to  the  most  simple  and 
commodious  form  it  is  capable  of;  and  that,  either  by 
bringing  it  to  its  least  terms,  or  all  the  members  thereofj 
if  it  be  compounded,  to  the  same  denomination. 


45  Reduction  of 

A  fraction  is  reduced  to  its  least  terms  by  dimding^ 
both  the  numerator  and  denominator  by  the  greatest  com- 
mon divisor » 

nh 

Thus,  --,  by  dividing  by  b^  is  reduced  to  - : 
be  c 

And  — 77->  by  dividing  by  ab^  is  reduced  to  ~  : 

abb  0^7  ^ 

Moreover, will  he  reduced  to  — ,  or  4^di 

Sab  1 ' 

And M=.  will  be  reduced  to  — 

72a^xWxy  6a 

Thus,  also,  — ^ — ^- ,  by  dividing    every  term  of 

the  numerator  and  denominator    by  2a,    is  reduced  to 
6a — ^b 

2a      ' 

^^^  6^^V:iTalF^ ^    ^^  ^^^^^^^S  every  terni 

by  2ax.  is  reduced  to : 

Sa-^2x 

Lastly,  !-~ 1- —T — ,   by  dividm^  both  the 

•^'  a^+3ab  +  2b^        ^      ^  ^ 

numerator  and  denominator    by  tUe   compound  divisor 

,    .        .        .^a^  +2ab+bb 

a  +  b.is  reduced  to • -I — . 

^   '  a  +  2b 

But  the  compound  divisors,  whereby  a  fraction  can 
sometimes  be  reduced  to  lower  terms,  are  not  so  easily 
discovered  as  its  simple  ones  ;  for  which  reason  it  may  not 
be  improper  to  lay  down  a  rule  for  finding  such  divisors. 

Firsts  divide  both  the  numerator  and  denominator  by  their 
greatest  simple  divisors^  and  then  the  quotients  one  by  the 
other  (as  is  taught  in  case  7,  section  5),  always  ob- 
serving to  make  that  the  divisor  which  is  of  the  least  di- 
"tnensions  ;  and^  \f(^^y  thing  remain^  divide  it  by  its  great- 
est sim.ple  divisor^  and  then  divide  the  last  compound  di-^ 


Fractional  ^antities.  47 

visor  by  the  quantity  thence  arising  ;  and  if  any  thing  yet 
remain^  divide  it  likewise  by  its  greatest  simple  divisor^  and 
the  last  compound  divisor  by  the  quantity  thence  arising; 
proceed  on  in  this  manner  till  nothing  remains;  so  shall  the 
last  divisor  exactly  divide  both  the  numerator  and  denomi- 
nator^ without  leaving  any  remainder. 

Note.  If,  after  you  have  divided  any  remainder  by  its 
simple  divisor,  you  can  discover  a  compound  one  which 
will  likewise  measure  the  same,  and  is  prime  to  the  divi- 
sor from  whence  that  remainder  arose,  it  will  be  conve- 
nient to  divide  also  thereby.  And,  if  in  any  case  it  should 
happen  that  the  first  term  of  the  divisor  does  not  exactly 
measure  that  of  the  dividend,  the  whole  dividend  may  be 
multiplied  by  any  quantity  that  shall  be  necessary  to  make 
the  operation  succeed. 

Ex.   1.     Let  it  be  required  to    reduce    the    fraction 

Sa'  +  lOa^'b-^SaH^      .     v     i         .  .  .     r  j 

to  Its  lowest  terms,    or  to  find 


a^b  +  2d'b'^  +  2ab^  +  ^^ 
the  greatest  common  measure  of  its  numerator  and  de- 
nominator. Here,  dividing  first  by  the  greatest  simple 
divisors,  5a^  and  ^,  we  have  a^  +  2ab  +  b^^  and  a^  + 
2a^b  -f  2a^2  ^  ^3 .  ^^^j  ^f  ^^  latter  of  these  be  divided  by 
the  former,  the  work  will  stand  thus  : 

a^  +  2a^b+   ab^ 
where  the  remainder  is  +    ab^  +  P  ;      which     be- 

ing divided  by  ^^,  its  greatest  simple  divisor,  gives 
a  +  b',  by  this  divide  a^  +  2ab  +  b^^  and  the  quotient 
will  come  out  a  +  b^  exactly ;  therefore  the  last  divisor, 
a  +  ^,  will  exactly  measure  both  quantities,  as  may  b^ 
proved  thus : 


a+b)  5a^  +  lOa^b  +  5a^o^  {Sa'^  +  5a^ 
5a^+    Sa'^b 


Sa'^b+Sa^b^ 
5^b  +  SaH^ 


48  Reduction  of 

a  +  F)  a^b  +  2aH^  +  2ab^  +  b^  (a^  +ab^+b'^ 

aH^+2ab^ 
a%^+    ab^ 


ab^  -f  b"^ 
ab^  4-  b'^ 
O         O 
In    both  which  cases  nothing  remains  ;     therefore   the 

fraction  eiven  will  he  reduced  to  — -. -. 

^  a^b-\-ab'^  A-b^ 

Ex.  2.     Let  it  be   proposed    to    reduce  the  fraction 
— -  to  its  lowest  terms:     and  then  the 


€? —  c^x  —  ax^  -f  x^ 
work  will  stand  as  follows  : 

.  ax^  4-  a:^)  a^  '+    O    +      0     -f    O  —  :\;*  («  +  x 
a^  —  a^x  — '    a^:\'^  +  ax^ 

a^x  +    a*A:^-— a^^—    x^ 
^3^_    (fx^'-^ax'^  +    x^ 
+  2a^x^  +0—2;^ 
a^  +  o— ..\^)  a^ — a^x  —  ax^  +x^  (a-^x 
a^  , —   o  — ax^ 

—  a^x  -f-    0    +  .x^ 

—  d^x  4-    0    -f  x^ 


0  0         0 

From  whence  it  appears  that  a^  ^  q  _^  ^,2^  ^^  ^2  _  ^2^ 
will  measure  both  a*^  —  x^  and  a^  —  a^x  -^ax^  +  x^  ;  and, 

by  dividing  thereby,  the  fraction  proposed  is  reduced  to 

^2  ^  ^2  ^ 


These  operations  are  founded  on  this  principle, 
that  xvhatever  quantity  measures  the  whole^  and  one  part 
of  another^  must  do  the  like  by  the  remaining  part.  For 
that  quantity,  whatever  it  be,  which  measures  both 
the  divisor  and  dividend,  in  the  first  example,  must 
evidently  measure  a*  +  2a^b  +  ab^  (being  a  multiple  of 


Fractional  ^antities.  49 

Example     3.     In     the     same     manner    the     fraction 

= ; —5 — ,  ^  3 Will   be    reduced    to 

. L- — •   See  the  process. 

X  —  Sa 
x^^^ax^ — ^a^x + 6a^)x^'--2>ax^ — Sa^^s-f  1  %a^x—^a\x — 2a 

— 2ax^-{'     0    +12a^x —  Sa'* 


remainder  — ^^(fi^jt^ —  4^^^;^' — :  ^a'^ ; 

which  divided  by  —  2a^^  gives  x"^  +  2(7a;  —  2a2  for  the 
next  divisor. 

x^  +  2ax  —  2a2)  a:^  —  ax^  —  ^a^x  +  6«3  (^  —  3a 
a:^  +  2ax^  —  2a^x 

—  tiax'^  —  Qa^x  -f  6a^ 

—  ^ax'^  —  6a^x  4-  6a^ 

0         o  o' 

;^2+2a;c— 2a2)Ar4— SaAT^ — %a^x'^+\^a^x — ^a^x"^ — SaX'\Aa^ 

x^-\-2ax^ — 2r^x^ 

^^5ax^ —  6a^x^  4. 1 8a^x 
*-^5ax^ — lOa^x'^  +  XOa^x 

+    4a^v2  4.    8c^x — 8a* 
4-    4a^x^+    8a^x — 8a^ 
0  0  0 

Now  if,  by  proceeding  in  this  manner,  no  compound 
divisor  can  be  found,  that  is,  if  the  last  remainder  be 
only  a  simple  quantity,  we  may  conclude  the  case  pro- 
posed does  not  admit  of  any,  but  is  already  in  its  lowest 


the  former)  :  whence,  by  the  principle  above  quoted, 
the  same  quantity,  as  it  measures  the  whole  dividend, 
must  also  measure  the  remaining  part  of  it,  ab^  -f-  b^ : 
but,  the  divisor,  we  are  in  quest  of  being  a  compound 
one,  we  may  cast  off  the  simple  divisor  ^^,  as  not  for 
our  purpose :  whence  a  +  b  appears  to  be  the  only  com- 
pound divisor  the  case  admits  of:  which,  therefore,  must 
be  the  common  measure  required,  if  the  example  pro- 
posed admits  of  any  such. 

II 


50  Reduction  of 

terms.      Thus,    for    instance,  if   the  fraction    proposed 

if^ere  to  be  — ;    it    is    plain    by 

or  +  ax  +  x^ 

inspection,  that  it  is  not  reducible  by  any  simple  divisor ; 
but  to  know  whether  it  may  not,  by  a  compound  one, 
I  proceed  as  above,  and  find  the  last  remainder  to  be  the 
simple  quantity  7xx:  whence  I  conclude  that  the  frac- 
tion is  already  in  its  lowest  terms. 

Another  observation  may  be  here  made,  in  relation 
to  fractions  that  have  in  them  more  than  two  different 
letters.  When  one  of  the  letters  rises  only  to  a  single 
dimension,  either  in  the  numerator  or  in  the  denomi- 
nator, it  will  be  best  to  divide  the  said  numerator  or  de- 
nominator (whichever  it  is)  into  two  parts,  so  that  the 
said  letter  may  be  found  in  every  term  of  the  one  part, 
and  be  totally  excluded  out  of  the  other ;  this  being 
done,  let  the  greatest  common  divisor  of  these  two  parts 
be  found ;  which  will,  evidently,  be  a  divisor  to  the 
whole,  and  by  which  the  division  of  the  other  quantity 
is  to  be  tried;  as  in  the  following  example,  where  the 

P       .         .         .     x^  +  ax^  4-  bx^  —  2a^x  +  bax  —  2ba^ 

traction  eiven  is  ■ ■ ■ ; • 

^  XX  —  bx  +  2ax  —  2ab 

Here  the  denominator  being  the  least  compounded,  and 

b  rising  therein  to  a  single  dimension  only,  I  divide  the 

same  into  the  parts  x^  -f-  2ax^  and  —  bx  —  2ab ;  which, 

by  inspection,  appear  to  be  equal  to  x  +  2a  x  ^-i  and 

X  -f  2a  X  —  b.     Therefore  .v  -f-  2a  is  a  divisor  to  both 

the    parts,    and    likewise    to    the    whole,    expressed    by 

X  +  2a  X  X  —  b  ;    so  that  one  of   these    two    factors, 

if  the  fraction  given    can  be    reduced  to  lower  terms, 

must    also    measure    the    numerator :    but    the    former 

will    be    found    to    succeed,    the    quotient    coming   out 

v2  —  ax  -j.  bx  —  ab^  exactly:  whence  the  fraction  it- 

-^  .         ,        ,       x^  —  ax  4-  bx  —  ab       i  .  i    . 
self  is  reduced  to ; ;  which  is  not  re- 

X 0 

ducible  farther,  by  x  —  b^  since  the  division  does  not 
terminate  without  a  remainder,  as  upon  trial  will  be 
found. 


Fjractional  ^antities*  51 

Having  insisted  largely  on  the  reduction  of  fractions  to 
their  least  terms,-  we  now  come  to  consider  their  reduc- 
tion to  the  same  denominator. 

Fractions  are  reduced  to  the  same  denominator  by  multi- 
plying the  numerator  of  each  into  all  the  deno7ninators^  ex- 
cept its  own^  for  a  nexu  corresponding  numerator^  and  all 
the  denominators  continually  together^  for  a  common  deno- 
7ninator. 

Thus,  ---and  —-will  be  reduced  to -— and  —-; 
b  d  bd  bd 

a      c  ^    e  adf    cbf  ,   bde 

_,  _,  and  -,  to  ^,  ^,   and    _; 

J  2ax         J    5bx         6a^x         ,    Sbxcd  , 

and  — -,  and  ,  to -,  .and  r  ;     and 

cd  3a        2acd  Sacd 

so  of  others. 

But  when  the  denominators  have  a  common  divisor, 
the  operation  will  be  more  simple,  and  the  conclusion 
neater,  if,  instead  of  multiplying  the  terms  of  each 
fraction  by  the  denominator  of  the  other,  you  only 
multiply  by  that  part  which  arises  by  dividing  by  the 
common  divisor.  As,  if  there  were  proposed  the  frac- 
tions — r  and  — -;  then,  the  denominators  havincr  the 
ad  cd  '    ° 

factor  d  common  to  both,  I  multiply  by  the  remaining 
factors    a   <ind   c  j    whence    the    two    fractions    will    be 

reduced  to  — ,  and  — -  (where    d  remains    as    before, 

acd  acd  ^ 

nothing  having  been    done    therewith).     By   the    same 

method    — —-    and     ■ — r—  are    reduced  to    --^    and 

Sabc  Aabd  *ZOabcd 

S5bcx^  ,     6aVax         ,     7c\^aa  -f  xx  ISa^Vax 

— — - — ■:    and    — ; —    and ; ,    to — 

20abcd        5bc  Sab        '  15abc 

,   SSc^Vaa  -f-  xx 

and     ■ . 

ISabc 

But,  as  has  been  before  hinted,  the  principal  use  of 
this  sort  of  reduction  is  to  transform  compound  quan- 
tities  to  the   most   commodious    forms  of   expression  j 


52  Reduction  of 

which,  for  the  general  part,  are  more  easily  managed 
(whether  they  are  to  be  added,  subtracted,  multiplied,  or 
divided)  when  all  their  members  are  brought  to  the  same 
denomination. 

Thus  the  compound    quantity  -5- J-— r  will  be  trans- 

0        a 

c         A    ^     da        he  ad  4-hc      f,       .     .  .  - 

lormed   to  ---  +  --  or  to  ^^ — ;   for   it   is   evident, 

ba        bd  bd 

that  the  quotient  which  arises  by  dividing  the  whole,  is 
equal  to  the,  quotients  of  all  the  parts,  by  the  same  di- 
visor. 

T     .1  ,.^    a        c    .  ad   —    he 

ixi  the  same  manner  will  --  —  —-  be  =  r-,— *  ; 

b         d  bd 

,  ^xy       Zx         c  _j_  2xybd  -f-  ISaaxd  *—  5aabc  ^ 
5aa        b         d  ^  Saabd  ^ 

,       2dx       /         2dx       b      .„  ,  2dx  -f  ab 

'  also,— 1-  /;,  or H  -  will  be  =  — — -i : 

a  a         \  a 

,    "lab  2ab  -^  aa  —  ah       ah  A-  aa 

and  J  +  a  =  ZL — =  — Z_.: 

a — b  a  —  b  a  —  b 

So  likewise,  by  reduction, u —  2a 

'     ^  '    a  —  x^  a-^x 


will  be  =  ^^  '^  a-i-x  +a^  X  a  —  ■  —  2a  X  a  +  ^"^^  X  a  —  x 

a>-^  X  X  ^  +x    « 

2a:^^  t^y/Z:.       10a  —  5x 

=     2 ^;  and  ilf^+--=rr = 

^^         ^'  a  Vxij  +  a 

5x1/  +  5aV*\y  4-  lOrt^  —  Sax         ,        .        , 

— " s^= ;  and  so  m  other  cases. 

aVxy  +  a^ 


The  reason  of  the  two  kinds  of  reduction  hitherto 
explained,  is  grounded  on  this  obvious  principle,  that 
the  equimultiples,  or  like  parts  of  quantities,  are  in  the 
same  ratio  to  each  other,  as  the  quantities  themselves  ;  or, 
that  the  quotient  which  arises  by  dividing  one  quantity 
by  another,  is  the  same  that  arises  by  dividing  any  part  or 
multiple  of  the  former,  by  the  like  part  or  multiple  of  the 


Fractional  ^lantities.  53 

Besides  these,  there  are  yet  two  other  sorts  of  reduc- 
tion which  authors  have  treated  of  under  the  head  of 
fractions  ;  which  are,  the  reducing  of  a  whole  quantity  to 
an  equivalent  fraction  of  a  given  denomination^  and  a  com* 
pound  fraction  to  a  simple  one  of  the  same  value.  Nei- 
ther of  these,  indeed,  are  of  any  great  use  in  the  sokition 
of  problems ;  however,  it  might  be  improper  to  leave  them 
entirely  untouched. 

1°.  A  whole  quantity  is  reduced  to  an  equivalent  frac- 
tion by  multiplying  it  by  the  given  denominator^  and  writ- 
ing the  midtiplier  underneath  the  product^  with  a  line  be- 
tween them. 

Thus  the  quantity  tf,  reduced  to  the  denominator  by 

will  be  — ,  and  'the  quantity  c  +  d^  to  the  denominator 

7      Mil     a-j-b  X  c  +  d       ac  -{-be  -^ad-^bd 

a  +  h.  will  be  — -n^—t  or — • 

'  a-^b        '  a  +  b 

2°.  A  compound  fraction  is  reduced  to  a  simple  one  of 
the  same  value  by  multiplying  the  numerators  together  for 
a  new  numerator  ^  and  the  denominators  together  for  a  nexu 
denominator.  .  • 

But  by  a  compound  fraction  here,  we  are  not  to  un- 


latter:  for,  in  reducing  to  the  lowest  terms,  it  is  plain, 
that,  instead  of  the  whole  numerator  and  denominator, 
we  only  take  that  part  of  each  which  is  defined  by  the 
greatest  common  measure ;  whereas,  in  reduction  to 
the  same  denominator,  we,  on  the  contrary,  make  use  of 
equimultiples  of  those  quantities  ;  since,  in  multiplying 
any  numerator  into  all  the  denominators,  except  its  own, 
we  multiply  it  by  the  very  same  quantities  by  which  its 
denominator  is  multiplied. 

The    rule     for     reducing    a    compound     fraction    to 
a    simple    one,    may  be    explained  thus.       It    is    plain 

c  1 

that  the    part  of  —  defined    by  -~,    which   arises   by 
a  b 

dividing   by  ^,   will  be  equal  to  ~    (the  divisor  here 

bd 


o4  Reduction  of 

derstand  one  consisting  of  several  tenns,  connected  toge- 
ther by  the  signs  +  and  —  (which  i^  the  general  definition 
of  a  compound  quantity),  but  such  a  one  as  expresses  a 
given  part  of  some  other  fraction. 

Thus  ^  of  -^  will    be    equal    to    ^,    and    the    ^ 
o  5  15  b 

part  of  ~-  will  be  =  7-/ 
a  bd 


Of  the  Reduction  qf  Radical  Quantities. 

The  reduction  of  surd  quantities,  like  that  of  fractions^ 
may  be  either  to  the  least  terms,  or  to  the  same  denomi- 
nation. 

A  radical  quantity  is  reduced  to  its  least  terms  by  resolv- 
ing it  into  factors^  and  extracting  the  root  of  that  which  is 
rationaU 

Thus,  V28  is  reduced  to  V4  x  VT";  which,  by 
extracting  the  square  root  of  4,  becomes  2\/  7 ;  also 
\^c^b  is  reduced  to  Va^  X  V  <^ ;  which,  by  extracting 
the  root  of   a^,    becomes  aV  b :     likewise,    K/a^b'^c^^  or 

1"^    is    reduced    to    s-cv'b^    x    V^c^    or    obK^bc^ : 

1 4a^x  —  4a^x^    .  -        ,  J  4a^ 

'->    \ 7~i ^^5    reduced    to    y— ^    x 

•    onH  ▲  /  ^_ 

Slb''c—'162b'x 


a''o'*c 
moreover. 


fax  —  x^         2a         lax  —  x^  ,'^ll6a^x'^+    16ttV 


c 
being  b  times  as  great) ;    therefore  the  part  of  -— •  de- 
fined by  --,  being  a  times  as  great  as  that  defined  by  -7-, 

c  ac 

must  be  truly  expressed  by  —  X  «,  or  its  equal  ~ ;  as  was 

to  be  shown. 


Radical  ^lantities.  55 


reduced    to    ^J—-  x  Sj-TZZZ 


a^x^  ^ax 

is    reduced    to    ^__--   x  V     2    '   ../.">    ^^'    ":;r    X 

4^/f?  -i-  ^  ^  ^^^  gQ  q£  ^^y  other:  all  which  is 
V   c^  —  2bx  \ 

evident  from  case  4  of  multiplication,  and  case  3  of  invo- 
lution. But  it  is  to  be  observed,  that,  in  resolving  any 
expression  in  this  manner,  the  factor  out  of  which  the 
root  is  to  be  extracted,  is  always  to  be  taken  the  greatest 
the  case  will  admit  of.  It  also  may  be  proper  to  take  no- 
tice, that  this  kind  of  reduction  is  chiefly  useful  in  the 
addition  and  subtraction  of  surd  quantities,  and  in  uniting 
the  terms  of  compound  expressions  that  are  commensurable 
to  each  other,  where  the  irrational  part,  or  factor,  after 
reduction,  is  the  same  in  each  term. 

Thus,  VI8  -f^Vsi  is  reduced  to_3V/2  +  4^2",  or 
rV2';  and  Vsf  +  \^50f  —  ^72^  is  reduced  to 
2aV"2"  +  5«VT  —  6aV  2  =  a>/  2.     Moreover,  by  re- 

.      .             }l2a^x     ,        flScFx    ,                           /48a^ 
duction,    V +   \] — - —    becomes  =   \J 4. 

'     ^      5  >      4  ^     20       ^ 

hsa^x        .      Hx   .    ^      f^       ^      I3x 

^    20  ^20  ^      ^  20  ^20 

And  SaS/4a^x^  +  Sx"^  -f  3xV9a  -f-  ISa^x^  becomes 
eaxVa^  +  2;c2  ^  9axVa^  +  2x^  ==  XSaxVa"  -f  2a,^ 

Surd  quantities^  under  different  radical  signs  ^  arc  reduced 
to  the  same  radical  sigiiy  by  reducing-  their  indices  to  the 
least  coinnion  denominator. 


3^,    reduced  to  the  same  radical  sign, 

will  become  a^  and  a^  (for  the  indices  are  here  i  and 
|-,  and  these  are  equivalent  to  f  and  |,  where  both  have 
the  same  denominator).     In  the  same  manner,  2"j^  and 

3J3   will  become  26"  and  3|^,  or  8|«   and  9|«.      And 

It  JL        '  ' 

universally^    A"*    and    B^     will,    when  their   exponents 


o6  jReduction  of^  £sf*c. 


are  reduced  to  the  same   denomination,   become  *A\^ 
andll 


mp 


The  principal  use  of  this  sort  of  reduction  is,  when  quan- 
tities under  different  radical  signs  are  to  be  multiplied  or 
divided  by  each  other. 


That  the  reduction  of  a  radical  quantity  to  another  of  a 
different  denomination,  by  an  equal  multiplication  of  the 
terms  of  its  exponent,  makes  no  alteration  in  the  value  of 
the  quantity,  may  be  tlius  demonstrated. 

m 

Let  A  "  be  any  quantity  of  this  kind  ;  then,  the  terms 
of  its  exponent  being  equally  multiplied  by  any  number 

mr 

r,  I  say,  the  quantity  A  ^^  hence  arising,  is  equal  to  the 
given  one  A**. 

For,  if  X  be  assumed  ==  A"'* ,  or,  which  is  the  same 
thing,  if  the  value  of  x  be  such,  that  x'*''  =  A  ;  then  the  nth 
root  of  x^"^  being  x'^  (by  case  2  of  section  6)  and  the  nth 

root  of  A  being  A"    (by  notation)^  these  two  quantities 

JL 
x**  and  A  ^    must,  likewise,    be    equal    to    each  other : 
and,  if  they  be  both  raised  to  the  72th  power,  the  equa- 
lity will  still  continue ;  but  the  72th  power  of  the  former 
(x^)  is  =  x'^'^  (by  case  2  of  involution)  ;    and  the  mth. 

1  m 

power  of  the  latter  (A "  )  is  A"«   (^by  notation)  ;    there- 
fore  ;c"^^  is  =  K'' .        But,  x  being  =  A«^,  we  have 

mr  mr  m 

^mr^^nr^  ^^  notation;  and  consequently  A"''=  A'*  ; 
which  xvas  to  be  proved. 


Of  Equations.  57 

Thus,  VT,  multiplied  by  a/10,  or  T^,    into   lo]^^^ 

will  give  125]^  X   lOu"!^,  or  12500l^  :   also,  Vax  into 

\/a^Xy  or  ax\^  into  6?^;^1^,  will  give  02:^^1^  x  a^^v*"]^, 

or    a';v^^^ :      and    Vax^    divided    by    Va^x    will    give 

fl^vV^  1^  ;^  J^  ...  - 

ZZZTT-i  o^  — 1  •        Lastly,    2x^  multiplied   into  VZax^ 
a4;rf]6         ^         ^ 

will  give  Vl^  X  Va^x*,  or  V\2ax^. 


SECTION  IX. 

Of  Equations. 

AN  equation  is,  when  two  equal  quantities,  differently 
expressed,  are  compared  together,  by  means  of  the  sign 
=  placed  between  them. 

Thus,  8  —  2  =  6  is  an  equation,  expressing  the  equa- 
lity of  the  quantities  8  —  2,  and  6  :  and  ^  =  a  +  6  is  an 
equation,  showing  that  the  quantity  represented  by  x  is 
equal  to  the  sum  of  the  two  quantities  represented  by  a 
and  b. 

Equations  are  the  means  whereby  we  come  at  such 
conclusions  as  answer  the  conditions  of  a  problem  ;  where- 
in, from  the  quantities  given,  the  unknown  ones  are  deter- 
mined ;  and  this  is  called  the  resolution  or  reduction  of 
equations. 

Reduction  pf  Single  Equations. 

Single  equations  are  such  as  contain  only  one  un- 
known quantity;  which,  before  that  quantity  can  be 
discovered,  must  be  so  ordered  and  transformed,  by  the 
addition,  subtraction,  multiplication,  or  division,  ^c* 
of  equal  quantities,  that  a  just  equality  between  the  two 
parts  thereof  may  be  still  preserved,  and  that  there  may 
result,  at  last,  an  equation,  wherein  the  unknown  quaiv 

I 


58  Of  Equatiorui. 

tity  stands  alone  on  one  side,  and  all  the  known  ones  on 
the  other.  But,  though  this  method  of  ordering  an  equa- 
tion is  grounded  upon  self-evident  principles,  yet  the  ope- 
rations are  sometimes  a  little  difficult  to  manage  in  the 
best  manner  -,  for  which  reason  the  following  rules  are 
subjoined. 

1  °.  Any  term  of  an  equation  may  be  transposed  to  the 
coritrary  side^  if  its  sign  be  changed^. 

Thus,  if  a:  +  6  =  16  ;    then  will  .v  =  16  —  6,  that  is. 

And,  li  X  —  4  =  8;  then  will  .v  =  8  -f  4,  or  x  =  12  ; 

Also,  if  3y  =  2;c  -f  24  ;  then  will  Zx  —  2x  :=.  24,  that 
is,  AT  =  24  : 

Again,  \^  5x  —  8  =  3x  +  20  ;  then  wall  Sx  '—*  ^x  ■=. 
20  +  8,  or  2x  =  28  : 

Lastly,  if  ax  +  bx  —  c  •\-  d  —  ex  =f —  g  -^hx  —  kx  ; 
then,  by  transposition,  ^ax  +  bx  —  ex  —  hx  +  kx  z=zf-^ 
g  +  c  —  d ;  where  all  the  terms  affected  by  x  (the  un- 
known quantity)  stand  now  on  the  same  side  of  the  equa- 
tion. 

2°.  If  there  be  a?iy  quantity  by  which  all  the  terms  of  the 
equation  are  multiplied^  let  them  all  be  divided  by  that 
quantity  ;  but  if  all  of  them  be  divided  by  any  quantity^  let 
the  common  divisor  be  cast  axvay. 

Thus,  the  equation  yza:  zzl  ab  is  reduced  to  x  =  b\ 
also,  lO.v  =  70  (or  10  X  -v  =  10  X  /)  is  reduced  to 
X  =  7 ;     and  x^  =  ax^  +  bx^  is  reduced  x  -=.  a  -{•  b'. 


^  The  reason  of  this  rule  is  extremely  evident ;  since 
transposing  a  quantity  thus  is  nothing  more  than  sub- 
tracting or  adding  it  on  both  sides  of  the  equation,  ac- 
cording as  the  sign  thereof  is  positive  or  negative.  Thus, 
in  the  equation  ^  +  6  =  16  (which,  by  transposition,  be- 
comes Ar  =  16 — ^6==10),  the  number  6  is  subtracted  from 
both  sides  ;  and  in  the  equation  x  —  4  =  8  (which,  by 
transposition,  becomes  a;  =  8  +  4  =  12),  the  number  4  is 
added  on  both  sides. 


Of  Equations*  59 

X         h   , 
Moreover  (bij  the  latter  part  of  the  rule)^  —  =  —  is  re- 

ax^       abx^     '  ac\ 

duced  to  X  =  d  i    and  = — ,  to  ax^  =  abx^ 

c  c 

—  acx^ ;   which,  if  the  whole  be  divided  by  ax^^  will  be 

farther  reduced  to  a:  =  ^  —  c. 

3°.  If  there  be  irreducible  fractions^  let  the  whole  equa- 
tion be  multiplied  by  the  product  of  all  their  denominators ^ 
or^  which  is  the  same  things  let  the  7iumerator  of  every  terin 
in  the  equation  be  multiplied  by  all  the  denominators^  except 
its  own^  supposing  such  terms  (if  any  there  be^  that  stand 
7vithout  a  denominator^  to  have  a  unit  subscribed*     / 

XX 

Thus,  the  equation  x  +  —  +  —  =11,  is  re^- 
duced  to  e>x  +  Zx  +2x  =  66  ;  and  x  +  ^^-t-  =  12  + 
^  ~  ^,  to  40x  +  Sa;-  +  16  =  480  +  5x  —  15  :     so, 

o 

T1       •  X        X  +  b   ,      *   ^        , 

likewise,  a = is  reduced  to  a^c  —  ex  =z  ax 

a  c 

ax  ex 

+  ab  ;    and f-  a  =  —-  to  cd^x  +  a^b  +  obx  =  acx 

a  +  x  b 

+  cx^. 

■  4"^.  If  ill  your  equation^  there  be  an  irreducible  surd^ 
wherein  the  unknown  quantity  enters^  let  all  the  other  terms 
be  transposed  to  the  contrary  side  (by  rule  1)  ;  and  then^ 
if  both  sides  be  involved  to  the  power  denominated  by  the 
surd^  an  equation  xvill  arise  free  from  radical  quantities  ;  un- 
less there  happen  to  be  more  surds  than  one^  in  which  case 
the  operation  is  to  he  repeated. 

Thus,  \^  X  +  6  =  10,  by  transposition,  becomes  V  x 
(=10  —  6)  =  4  ;  which,  by  squaring  both  sides,  gives 
X  =16.  -  

So,  likewise,  \^aa  -^  xx  —  c  ■=.  x^  becomes  Vaa  +  xx 
■==z  c  +  X  'y  which,  squared,  ,gives  aa  +  xx  =  cc  +  2cx 
f  wv,  o]*  aa* —  cc  =  9.cx  {by  rule  1).      The  reasons  of 


60  OfEquattons^ 

this,  as  well  as  of  the  two  preceding  rules,  depend  on  self- 
evident  principles  :  for,  when  the  equal  quantities,  on  each 
side  of  an  equation,  are  multiplied  or  divided  by  the  same, 
or  by  equal  quantities,  or  raised  to  equal  powers,  the  quan- 
tities resulting  must  necessarily  be  equal. 

5°.  Having'^  by  the  preceding  rules  {if  there  he  occasion)^ 
cleared  your  equation  of  fractional  and  radical  quantities^ 
and  so  ordered  it^  by  transposition^  that  all  the  terms^  where* 
in  the  unkno-wn  quantity  is  founds  may  stand  on  the  same 
side  thereof  let  the  whole  be  divided  by  the  coefficient^  or  the 
sum  of  the  coefficients^  of  the  highest  power  of  the  saidwi- 
known  quantity :  and  then,  if  your  equation  be  a  simple 
one  (that  is,  if  the  first  power,  or  the  quantity  itself,  be 
only  concerned),  the  work  is  at  an  end  ;  but  if  it  be  a  qua- 
dratic or  cubic  one,  &Pc.  something  further  remains  to  be 
done  ;  and  recourse  must  be  had  to  the  particular  me- 
thods for  resolving  these  kinds  of  equations,  hereafter  to 
be  considered  in  their  proper  place. 

I  shall  here  subjoin  a  few  examples  for  the  learner's 
exercise,  wherein  all  the  foregoing  rules  occur  promise 
cuously. 

Ex.  1.  Let  5x  —  16  =  3;^  +  12:  then  (%  rule  1) 
5x  —  3x  =  12  +  16,  or  2x  =  28  :    whence  {by  rule  5) 

.V  =  —  =  14. 
2 

Ex.  2.  Let  20  —  3;^  —  8  =  60  —  7x:  then,  —  3;^  -f-  Tx 

=  60  —  20  -f-  8,  that  is,  4a:  =  48  ;   and  consequently  x  z= 

4 
Ex.  3.  Let  ax  —  b  :=zcx  +  d;    then  ax  **^  ex  :=:  d  +  b^ 

or  a— ^  XX  =  d  +  b  ;    and  therefore  x  =  (by 

rule  5*). 

Ex.  4.  Let  6x^  —  20x  :=zl6x  +2x^t  then,  dividing  by 
2x  {according  to  rule  2)  we  have  3x  —  10  =  8  -f-  a:  ;  whence 
3;^-  -— .  a;  =  8  +  10,  that  is,  2;^  =  18  :  and  therefore  x  =^- 


o 


Of  Equations,.  61 

Ex*  5.  Let  oax^  — -  ahx^  =  ax^  ^f-  "lacx^ :  here,  dividing 

the  whole  by  ax^^  we  have  Zx  -^b  =  .v  +  2c ;    therefore 

7        1  2c +  ^ 

2j^  =  2c  +  ^,  and  .r  =  — - — . 

^  -^ 

^a;.  6.  Let  ^  +—  =  21  :    then  (%  rw/e  3)  4.v  +  3;^- 
3        4 

252  ' 

=  252  ;  and  therefore  x  =— —  =  36. 

^;,.  7.  Let  ^±i  +  ^i±^  =  16  ^  ^±1 :  then  12:^  +  12 
2  3  4 

+  8:v  +  16  =  384  —  ^x  —  18  :    whence  26.v  =  338,  and 

338       ,^ 

A^  = =  13. 

26 

Ex.  8.  Let  a =  c  :  then  ax  "--^bb  =.  ex  ;    whence 

X 

ax  *—  c^*  =  bb^  and  x  =:  ■ 


a  —  c 


JE:v.  9.  Let h  -7-  H =  ^-    ^^^^5    ^^-^  +  ^^^  + 

a         b         c 


abx  =  rt/!?c^,  or  be  +  ac  +  ab  x  x:  =  abed;  and  consequent- 
ly ;v  =  -■ (bt/  rule  5). 

•^  bc  +  ac  +  ab^  ^  ' 


axP'  4.  rtc^ 


i::v.  10.    Let  ax  +  b^  ^ "^ :    then,  ax  +  h^  x 

a+  X 


n  +  xz=  ax^  +  ac^j  that  is,  a^x  +  ab^  +  ^a:^  +  b^v  = 
r/;c2  +  ^r^  ;    whence  a^x  +  <2X^  +  b^x  —  ax^  =  ac^  -— 
a^2,    or  rt^A"  +  b^x  =  ^c^  —  ab^  ;     and  therefore  a:  =:: 
ac^  —  ab^ 
aa  +  bb* 

Ex.  11.    Let 1 =  1  :    then  ax  +  ab  +  bx 

a  +  ;f        X 

•=.  ax  -^  XX  ;  whence  —  xx  •\-  bx,  =  ■ —  abi    which,  by 

changing  all  the  signs  (in  order  that  the  highest  power 

of  X  may  be  positive),  gives  xx  —  bx  z=l  ab.      But  the 

same  conclusion  may  be  otherwise  brought  out,  by  first 

changing  the  sides  of  the  equation  ax  +  ab  +  bx  =  nx 


^^  OfEquatiom. 

+  XX ;  which  thereby  becoming  ax  +  xx=:ax  +  ab4.  l>x 
we  thence  get  xx  ~i>x  =  ab,  the  same  as  before. 

Ex.   12.      Let  :^  +  12  =   ir:     then    ^  -  ^ 

and  V/5X  =  15;    whence  {by  rule  4),  5;.  =  225,  and 

therefore  x  =  — £  =  45. 
5 

iix.  13.  Let  VlTJ^  =  2  +V/^  :  then  (^y  rw/e  4) 
12+^_=4  +  4Vx  +  x;  whence,_by  transposition, 
8=  4V^  ;    and,  by  division,  2  =  V:^  ;    consequently, 

E..  14.    Let.v  +  V^^-+.-  =  ;;;-==.      Here  {by 

rulej>)_x  X  Va^+x^  X  a^  +  x^  =  2a^  ;    whence  x  x 

V'a^  +  x^  =  a^  —  x^^  and  x^  X  a^  +  x^  =z  a"^ 2aV 

+  x"^  (by  rule  4),  that  is,  a^x^  +  x"^  =z  a^ q^x^  +  x"^  • 

therefore  Za'x^  =  a\  and  ^c^  =  -fl  =  2!. 

Ex,   15.     Let  V  X    +  Va  +  X  =  —.      Then 

Va  +  X 

Vax  +  XX  +  a  +  X  z=z  2a^  or  Vax  +xxz:za  —  x;  whence 

ax  +  XX  :=  a^  —  2ax  +  x^.  and  a;  =  —  =  ~ 

3a       3* 

Ex.  16.  Let  v'^s  —  a^  =z  X  —  c:  then,  by  cubing 
both  sides,  x^  —  a^  =  x^  —  Sc;^^  j.  3c^x  —  c^  ;  whence 

Scx^  —  Sc^x  =za^  —  c\  and  x^ —  ex  z:z  —  ^tL^hy  divid- 

oc        3      "^ 
ing  the  whole  by  oc, 

Ex.  17.  Let  v^«a  4-  AT^  =;=  \/^4  _j^  ^4  .  ^hen,  by  raising 
both  sides  to  the  fourth  power  we  have  aa  +  xx"]^  =  ^^  -f 
a'%  that  is,  .^^  +  2aV  -f  at^  =:  Z^'*  +  at^  ;  and  consequently 


1^/2 


2^2        :2aa       ^' 


Of  Equations*  63 

Ex..  1 8.  Let  X  =  \a^  +  x  Vbb  +  xx  -r-  a.  Here  .v  +  a 
=r  \a^  +  X  s/bb  -f  XX  ;  which,  squared,  gives^  jv^  +  2a.v 
+  fl2  --.  ^  ^  ^  \/}jl)  j^  xx^  or  AT^  +  2c7Ar  =  :^V/>^  +  ivA; ;  di- 
vide by  ^,  so  shall  x  +  2a  =  \/bb  +  x:^  ;  this  squared  again 
gives  a:^  +  4ax  +  4a^  =  bb  +  xx ;  whence  4^wr  =  ^^  —  4a^, 

and  therefore  x  =. a. 

4a 

Of  the  Extermination  of  Unknown  Qtcantities,  or  the 
reduction  of  two  or  more  equations  to  a  single  one. 

It  has  been  shown  above,  how  to  manage  a  single 
equation ;  but  it  often  happens,  that,  in  the  solution  of 
the  same  problem,  two,  three,  or  more  equations  are  con- 
cerned, and  as  many  unknovvn  quantities  occur  pro- 
miscuously in  each  of  them  ;  which  equations,  before 
any  one  of  those  quantities  can  be  known,  must  be  re- 
duced into  one,  or  so  ordered  and  connected,  that,  from 
thence,  a  new  equation  may  at  length  arise,  affected  with 
only  one  unknown  quantity.  This,  in  most  cases,  may  be 
performed  various  ways,  but  the  following  are  the  most 
general. 

1°.  Observe  xvhich  of  all  your  tmknoxvn  quantities  is  the 
least  involved^  and  let  the  value  of  that  quantity  be  found 
in  each  equation  {by  the  methods  already  explained)^  looking' 
upon  all  the  rest  as  knoxun  ;  let  the  values  thus  found  be  put 
equal  to  each  other  {for  they  are  equal^  because  they  all  ex- 
press the  same  thing)  ;  whence  fiexv  equations  -w'llL arise ^ 
out  of  which  that  quantity  will  be  totally  excluded;  xvith 
these  new  equations  the  operation  may  be  repeated^  and  the 
unknown  quantities  exterminated^  one  by  one^  till^  at  last^ 
you  come  to  an  equation  containing  only  one  unknoxvn  quan- 
tity. ^    ' 

2  .  Or,  let  the  value  of  the  unknoxvn  quantity^  xvhich 
you  would  first  exterminate^  he  found  in  that  equation 
xvherem  it  is  the  least  involved^  considering  all  the  other 
quantities  as  known;  and  let  this  value^  and  its  powers^  be 
substituted  for  that  quantity  audits  respective  powers  in  the 


6i  Of  Equattom* 

other  equations  ;  andy  -with  the  nexv  equations  thus  arising^ 
repeat  the  operation^  till  you  have  only  ane  unknown  quan- 
tity and  one  equation. 

3°.  Or,  lastly^  let  the  given  equations  be  multiplied  or 
divided  by  such  numbers  or  qua7itities^  whether  known  or 
unknown^  that  the  term  which  involves  the  highest  power  of 
the  unknown  quantity  to  be  exterminated^  may  he  the  same 
in  each  equation  ;  and  then  by  adding^  or  subtracting  the 
equations^  as  occasion  shall  require^  that  term  will  vanish^ 
and  a  new  equation  emerge^  wherein  the  number  of  dimen- 
sions ( if  not  the  number  of  unknown  quantities^  will  be  di- 
minished. 

But  the  use  of  the  difFereixt  methods  here  laid  down 
will  be  more  clearly  understood  by  help  of  a  few  exam- 
ples. 

EXAMPLE  I. 

Let  there  be  given  the  equations  x  +  y  -=.  12,  and  5x  + 
Sy  ■=  50 ;  to  findx  and  y. 

According'  to  the  frst  method^  by  transposing  2/  and  3y, 
we  get  .v=  12  —  z/,  and  5x  =^  50 —  3z/ ;   from  the  last 

of  which  equations,  x  =  — Zl^.      Now,  by  equating 

,  r  1_  ...  50—32/ 

these  two  values  01  x^  we  have  12  — '  ^  = — -l 

and  therefore  60  —  52/  =  50  —  3z/ :    from  which  y  is 

given  =  —  =  5  ;  and  :\^  (=  12  —  z/  =  12  —  5)  =  7- 

According  to  the  second  method^  x  being,  by  the  first 
equation,  =  12  —  z/,  this  value  must  therefore  be  substi- 
tuted in  the  second,  that  is,  60  —  5y  must  be  written  in  the 
room  of  its  equal  5x  ;  whence  will  be  had  60  —  5y  +  2y 

=  50 :  and  from  thence  z/  =  —  =  5^as  before. 

But^  according  to  the  third  method^  having  multiplied 
the  first  equation  by  5,  it  will  stand  thus,  5:r  +  5z/  =  60  ^ 
from  whence  subtracting  the  2d  equation,  5x  +  3z/  =  50, 
there  remains  -  -  ?^  ^  ^  9  > 

whence  y  =  5?  still  the  same  as  before^ 


Of  Equations,  65 

The  first  of  these  three  ways  is  much  used  by  some 
authors  ;  but  the  last  of  them  is,  for  the  general  part, 
the  most  easy  and  expeditious  in  practice,  and  is,  for 
that  reason,  chiefly  regarded  in  the  subsequent  exam- 
ples. 

EXAMPLE  IL 

Let  i'^^-^^y-  ^l^ 

Here  the  second  equation  being  multiplied  by  4,  in  or- , 
der  that  the  coefficients  of  y  in  both  equations  may  be  the 
same,  we  have  12x —  8y  =  80. 

Let  this  equation  and  the  first  be  now  added  to- 
gether ;     whence  y  will  be  exterminated,  there  coming 

204 
out  17x  :=.  204 ;  from  which  x  =  - —  =  12  :  therefore, 

17 

,       ,      ^                  .             ,      124^  — 5x       124  —  60       64. 
by  the  first  equation,  y  (=  ^ —  =  — . ->  =  — ) 

=  8. 

EXAMPLE  in. 

^.  i    5x  —  3v=    90 

Given.   <    ^      ^    J'       .^^ 
(^    2:v  +  5z/  =  160. 

Here  multiplying  the  first  equation  by  2,  and  the  second 

by  5,  in  order  that  the  coefficient  of  x  may  be  the  same 

in  both,  there  arises 

\0x —    6z/=  180 

10^  +  25z/  =  800. 

By  subtracting  the  former  of  which  from  the  latter  we  havfc 

312/  =  620 :  hence  y  = =  20  j  and  so,  by  the  first 

oX 

,      90  4-3Z/       90  4-60. 
equation,  x  (= ^— ^  = — )  =  30. 

But  the  value  of  x  may  be  otherwise  found,  inde- 
pendent of  the  value  of  y ;  for,  by  multiplying  the  first 
equation  by  5,  and  the  second  by  3,  and  then  adding  them 
together,  y  will  be  exterminated,  and  you  will  get  25a;  -f- 

930 
^x  =  450  -I-  480 ;  whence  x  =  -;;--  =  30,  the  same  as  be- 

o  1 

Jbre, 

K 


66  Of  Equations. 

EXAMPLE  IV. 

[-+^  =  16 


Given.  <J 

I     X 


2         3 

y^^     cy 

L    5  9 

Here  our  equations,  cleared  of  fractions,  will  be 
^x  +  2y  =z  96 

9;^_52^=:90. 

And,  if  from  the  triple  of  the  former  the  latter  be  sub- 
tracted, we  shall  have  Gz/  +  5?/  =  288  —  90,  that  is,  ll?y 

96  —  22/. 
=  198  )  whence  ?/  =  18  j  and  x  (= — ^)  =  20. 

o 

EXAMPLE  V. 


Given.  < 


^_12  =  ^+8 
2  4 

5         :i  .4 


Here  4>;  —  96  =  2?/  +  64,  and 

12a;   +   122/    +    20.r  —  480  =   30z/  ~   l^x   +  1620; 

which,  contracted,  become 

Ax —    2z/ =    160,  and 

4i7x — 182/ =  2100:     from   the    last  of   which    subtract 

9  times  the  former;     so  shall  11a:  =  2100  —  1440  = 

4fX  — -  160 
660  ;  therefore  x  =  60,  and  y  (= z=z2x  —  80} 

=  40. 

EXAMPLE  VI. 


r  x  +  2/  =  i3^ 

Let  <    a;  +  2  =  14  >  ;  to  find  ^, «/,  and  ; 
t  2/  +2  =  15  J 


-  y  ■ 

By  subtracting  the  first  equation  from  the  second,  in 
order  to  exterminate  ac,  we  have  z  -—  2/  =  1  ;  to  which 
the  third  equation  being  added,  2/  will  likewise  be  exter- 
minated, there  coming  out  22  =  16,  or  2  =  8  ;  whence  y 
(^z^l)=i7'r  and:v(=13  — j/)  =  6. 


Qf  Equations. 


EXAMPLE  VIL 


Let 


2   ^   3   ^  4 

^       y       ^       A^ 

S         4^         5      . 

—  +  ^  +  —  =  38. 
L4  ^   5         6 


Here  the  given  equations,  cleared  of  fractions,  become 
12x  +    Sy  +    62  =  1488 
20;^  +15y  +  12zz=z  2820 
30;^  +  24z/  +  202  =  4560. 
Now,  to  exterminate  2,  let  the  second  of  these  equations 
be  subtracted  from  the  double  of  the  first ;   and  also  the 
triple  of   the  third  from  the  quintuple  of  the  second ; 
^hence  is  had 

4fX  +    y  =:  156 
10x  +  3y=z420: 

from  which  12^;  —  10;\r  =  468  —  420,  and  at  =  —  =24. 

2 

Therefore y  (=  156— 4;c)  =  60 ;  andz  (=il?iz:2^Ili2f ) 
=  120. 

EXAMPLE  VIIL 


r^  +  ioo=  y  + 

<y  +  100=  2x  +  \ 
\^z  +  100=  3;^ +  ' 


z 

Let  \y  +  100  =2x  +  2z 
Zx  +  32/. 

To  the  double  of  the  first  let  the  second  equation  be 
added  ;  so  shall  the  xes,  on  the  contrary  sides,  destroy  each 
other,  and  you  will  have  300  +  z/  =  2z/  +  42,  or  300  = 
y  +  42.  Moreover,  to  the  triple  of  the  first  let  the  third 
equation  be  added,  whence  will  be  had  2  +  400  =  6z/  +  32, 
or  400  =  6i/  +  22. 

Now,  if  from  the  double  of  this  last  equation  the 
former,  300  ■=  y  +  4<z^  be  subtracted,  there  will  come 

out  500  =  llj/;    and,  consequently,  y  =  —  =  45-/j ; 


63  Of  Equations. 

therefore  2;  (=  ^^ ~'^  =  75  — •  ^  =  75  —  11-^  = 

4  4        »  11'' 

eS-Z-j. ;  and  .r  (=  ^/  +  2;  —  100  =  109^V.  _  loo)  =  ^^. 

EXAMPLE  IX. 

Let  X  —  ?/  =  2,  and  xy-\-5x  —  6j/  =  1 20  ;  to  extermi- 
nate X. 

By  the  former  equation,  a:  =  z/  4. 2  ;  which  value  being 
substituted  in  the  latter  (according  to  the  second  general 
method^  it  becomes  ^/  +  2x^/  +  5x?/  +  2  —  6z/=  I2O5 
that  is,  2/2  +  2y  +  5y  +  10  —  6y  =  120,  or  2/^  +  2^  =  1 10. 

EXAMPLE  X. 

Let  there  be  given  x  +  y=za^  and  x^  •\' y^  :=.  h  ;  to  ex- 
terminate AT. 

Here,  by  the  first  equation,  x  ::rz  a  —  y  ,  and  therefore 
o^=:a  —  y'Y ;  which  value  being  written  in  the  other  equa- 
tion, w^e  have  a  —  y'Y'  +  y^  —  b^  that  is,  a^  —  2ay  +  y^  +  y^ 

==  b  ;  and  therefore  y^  —  ^y^=  ■ • 

EXAMPLE  XL 

Given    s  y  ^  ;     '^    ,  l         i  >  io  exterminate  y. 

Multiply  the  first  equation  by  f  and  the  second  by  ^, 

and  subtract  the  latter  product  from  the  former ;  whence 

you  will  have  bfx  —  agx  +  cfy  —  ahy  =  df —  ak  -,  which, 

,  ,.   .  .          .               df- — ak4-as^x — bfx 
by  transposition  and  division,  gives  y^-^ 7        i^— . 

Let  this  value  of  y  be  now  substituted  in  the  first  equa- 
tion, and  there  will  arise 
adfx — a^kx  -f-  a^gx"^ — abfx'^  -f  cdf —  cak  +  cagx  —  cbfx 

bx  -rz  d\    which,  multiplied  by  cf — ah^  and  contracted, 
gives  ag  — -  bfx  !^^  +  df — ■  ak  +  eg  >-^  bh  X  x  =■  ck '—  hd* 


Of  Equations.  69 

EXAMPLE  XII. 

Supposing  ax^  +bx-\'C'=  0,  and  fx^  +gx  +  A  =  0  ;  to 
exterminate  x. 

Proceeding  here  as  in  the  last  example,  we  have  fbx 

^fc  —  agx  —  <^/i  =  O ,   and,  from  thence,  x  =  7- J—. 

...  ah  — /?T       b  X  ah  — fc 

Whence,  by  substitution,  a  X  —  -  H Ti ' 

^  Jb  —  agJ         Jb  —  ag 

4.  c  =  0,     This,  by  uniting  the  two  last  terms,  and  divid- 

3        111             •        <2/2 — fcY      bh  —  eg      ^ 
mg  the  whole  by  a,  gives  ^    ■-  +  -7- =  O  ;  con- 

fb  —  agj     f^  —  «.§' 

sequently,  ah  — fc\^  +fb  —  ag  X  bh  —  c§*  =  0. 

After  the  same  manner  x  may  be  expunged  out  of  the 
equations  ax^  +  bx"^  -f  rx  -f-  <^  =  O,  and  fx^  -f  gx  +  A  =  0, 
&c.  But,  to  show  the  use  of  the  above  example,  sup- 
pose there  were  given  the  equations  x^  -{-yx  ^-^y^  =:  0, 
and  x^  +  3xy  —  10  =  O  :  then,  by  comparing  the  terms 
of  these  equations  with  those  of  the  general  ones,  ax^  + 
bx  +  c  =:0y  and  fx^  +  gx  +  h  =z  0  ;  we  have  «  =  1, 
b  =z  y^  c  =  —  2/%  /  =:  1,  g  z=:  Sy^  and  h  =  —  10; 
which  values  being  substituted  in  the  equation  ah—Jc]^ 
+fb  —  ag  X  bh  —  c^  =  0,  it  thence  becomes  —  10  +  yyy 
4  y  —  2>y  X  —  lOz/  -f  3i/2  =  0,  that  is,  100  —  202/^  +  y^ 
+  20?/^  —  62/"*  =  0 ;  or,  100  =  5//^;  whence  y  may  be 
found,  and  from  thence  the  value  of  x  also. 


SECTION  X. 

Of  Proportion. 


QUANTITIES  of  the  same  kind  may  be  compared 
together,  either  with  regard  to  their  differences,  or  ac- 
cording to  the  part  or  parts  that  one  is  of  the  other,  call- 
ed their  ratio.     The  comparison  of  quantities  according 


70  Of  Proportimu  ^ 

to  their  difFerences,  is  called  arithmetical;  but,  according 
to  their  ratios,  geometrical. 

When,  of  four  quantities,  2,  6, 12,  16,  the  difference  of 
the  first  and  second  is  equal  to  the  difference  of  the  third 
and  fourth,  those  quantities  are  said  to  be  in  arithmetical 
proportion.  But,  when  the  ratio  of  the  first  and  second 
is  the  same  with  that  of  the  third  and  fourth  (as  in  2,  6, 
10,  30),  then  the  quantities  are  said  to  be  in  geometrical 
proportion.  Moreover,  when  the  difference,  or  the  ratio, 
of  every  two  adjacent  terms  (as  well  of  the  second  and 
third,  as  of  the  first  and  second,  ££?c.)  is  the  same,  then  the 
proportion  is  said  to  be  continued:  thus,  2,  4,  6,  8,  &fc.  is 
a  continued  arithmetical  proportion  ;  and  2,  4,  8,  16,  ^c. 
a  continued  geometrical  one.  These  kinds  of  proportions 
are  also  called  progressions,  being  carried  on  according  to 
the  same  law  throughout. 

Arithmetical  Proportion. 

THEOREM  I. 

Of  any  four  quantities^  a,  Z»,  c,  </,  hi  arithmetical  pro- 
gression^^ the  sum  of  the  two  means  is  equal  to  the  sum  of 
the  two  extremes. 

For  since,  by  supposition,  ^ — a  is  =:  d — c,  therefore 
is  b  -^-c  =:d  +  ajhy  transposition. 

THEOREM  II. 

In  any  continued  arithmetical  progression  (5,  7,  9,  11, 
13,  15)  the  sum  of  the  txvo  extremes^  and  that  of  every  other 
txvo  terms  equally  distant  from  them^  are  equal. 

For  since,  by  the  nature  of  progresslonals,  the  second 
term  exceeds  the  first  by  just  as  much  as  its  corresponding 


^  Although,  in  the  comparison  of  quantities  according 
to  their  differences,  the  term  proportion  is  used ;  yet  the 
word  progression  is  frequently  substituted  in  its  room, 
and  is,  indeed,  more  proper ;  the  former  term  beihg,  in 
the  common  acceptation  of  it,  synonymous  with  ratio, 
which  is  only  used  in  the  other  kind  of  comparison. 


Of  Proportion,  71 

term,  the  last  but  one,  wants  of  the  last,  it  is  manifest, 
that,  when  these  corresponding  terms  are  added  together, 
the  excess  of  the  one  will  make  good  the  defect  of  the  other, 
and  so  their  sum  be  exactly  the  same  with  that  of  the  two 
extremes:  and  in  the  same  manner  it  Will  appear,  that 
the  sum  of  any  other  two  corresponding  terms  must  be 
equal  to  that  of  the  two  extremes. 

When  the  number  of  terms  is  odd,  as  in  the  progres- 
sion, 4,  7,  10,  13,  16,  then  the  sum  of  the  two  extremes 
being  double  to  the  middle  term,  or  mean,  the  sum  of  any 
other  two  terms,  equally  remote  from  the  extremes,  must 
likewise  be  double  to  the  mean. 

THEOREM  III. 

In  any  continued  arithmeUcal progression^  a^  a  -\-  d^  a  •\' 
2d^  a  +  3(3^,  a  -f-  4<^,  &f  c.  the  last^  or  greatest  term^  is  equal 
to  the  firsts  or  least ^  more  the  common  difference  of  the  terms 
drawn  into  the  number  of  all  the  terms  after  the  Jirst^  or 
into  the  xphole  number  of  terms ^  less  one* 

For,  since  every  term,  after  the  first,  exceeds  that  pre- 
ceding it,  by  the  common  difference,  it  is  plain  that  the 
last  must  exceed  the  first  by  as  many  times  the  common 
difference  as  there  are  terms  after  the  first ;  and  therefore 
must  be  equal  to  the  first,  and  the  common  difference  re- 
peated that  number  of  times. 

THEOREM  IV. 

The  sum  of  any  rank  or  series  of  quantities^  in  continued 
arithmetical  progression  (5,  7,  9,  11,  13,  15),  is  equal  to 
the  sum  of  the  two  extremes  multiplied  into  half  the  number 
of  terms. 

For,  because,  by  the  second  theorem,  the  sum  of  the 
two  extremes,  and  that  of  every  two  other  terms  equally 
remote  from  them  are  equal,  the  whole  series,  consist- 
ing of  half  as  many  such  equal  sums  as  there  are  terms, 
will  therefore  be  equal  to  the  sum  of  the  two  extremes 
repeated  half  as  many  times  as  there  are  terms.  The 
same  thing  also  holds  good,  when  the  number  of  terms  is 
odd,  as  in  the  series  8,  12, 16,  20,  24 ;  for  then  the  mean, 
or  middje  term,  being  equal  to  half  the  siun  of  any  two 


72  J         Of  Proportion. 

terms  equally  distant  from  it,  on  contrary  sides,  it  is  ob- 
vious that  the  value  of  the  whole  series  is  the  same,  as  if 
every  term  thereof  was  equal  to  the  mean,  and  there- 
fore is  equal  to  the  mean  (or  half  the  sum  of  the  two 
extremes)  multiplied  by  the  whole  number  of  terms ;  or 
to  the  whole  sum  of  the  extremes  multiplied  by  half  the 
number  of  terms. 

Geometrical  Proportion. 

THEOREM  I. 

If  four  quantities^  a,  ^,  c,  d  (2,  6,  5,  15),  be  in  geome- 
trical proportion^  the  product  of  the  two  means^  bc^  will  be 
equal  to  that  of  the  two  extremes^  ad. 

For,   since  the  ratio  of  a   yjo  b  (ur  the  part  which  a 

is  of  ^)  is  expressed  by  ---,  and  the  ratio  of  c  to  d^  in  like 

Co 

manner,  by  — -  ;  and  since,  by  supposition,  these  two  ratios 
d 

are  equal,  let  them  both  be  multiplied  by  bd^  and  the  pro- 

d  c 

ducts    ,  X  bd  and  --^xbd  will  likewise  be  equal ;  that  is, 
b  d 

--—  =  --— ,  or  ad  =  cb  (by  case  2,  sect.  4). 
b  d 

THEOREM  II. 

If  four  quantities^  c,  ^,  c,  d^  be  such  that  the  product  of 
two  of  them ^  ad^  is  equal  to  the  product  of  the  other  two^  bc^ 
then  are  these  quantities  proportional. 

For  since,  by  supposition,  the  products  ad  and  be  are 
equal,    let  both  be  divided    by  bd^    and    the    quotients 

^  (-^J  and  -1  (i-)  will  also  be  equal;  and  therefore 
bd   \b  I  bd  \  d I 

a\b  \  \  c  \  d. 

THEOREM  IIL 

If  four  quaiitities^  a,  ^,  c,  ^(2,  6,  5,  15),  be  proportion- 
al^ the  rectangle  of  the  means y  divided  by  either  extreme^ 
will  give  the  other  extreme. 


Of  Proportloiv.  73 

For,    by  the    second    theorem,  ad  =  be  (2  X   '^S  z=^ 
6  X  5),  whence,  dividing  both  sides  of  the  equation  by 

a  (2),  we  have  d  =z  —  (15  ==  — ^- — 1.      Hence,  if  the 

two  means  and  one  extreme  be  given,  the  other  extreme 
niay  be  found. 


THEOREM  IV. 

The  products  of  the  corresponding  terms  of  txvo  geome-^ 
irical  proportions  are  also  proportional. 

That  is,  \i  a  \  b  I  \  c  \  d^  and  e  \  f  \  :  g  :  h  ;    then  v/ill 
ae  :  If:  :  eg  :  dJu 

For  —-  =  —-,  and  -^  =  — ,  by  supposition  ;     whence 
b  d  J        f^ 

ti        e        e         p" 
^  X  "7  ~  "7  ^  "X'  ^y  equal  multiplication ;    and  conse- 

ae      csr 
quently  -j~-z=z^   (by  p.  18)  ;    that  is,  ae  \bf  \  :  eg  \  dh* 

Hence  it  follows,  that,  if  four  quantities  be  proportional, 
their  squares,  cubes,  &7'c.  will  likewise  be  proportional. 


THEOREM  V. 

tffour  quantities^  a^  b^  c,  d  (2,  6,  5,  15),  be  proportional^ 

f  1.  inversely,  b  :  a  i  ',  d  :  e  {6    :  2    :  :  15  :    5) 

j  2.  alternately,  ai  c:\  b:  d(Jl    :  5    :  :    6:15) 

^  I  3.  compoundedly,     a:a+b::c:c-^ I2    :  8    :  :    5  :  20) 

g   ]  4.  dividedly,  a:b — a::c:d—e  (2    :  4    :  :    5:10) 

^   j  5.  mixedly,    b+a:b — aiid-hcid-^c  (8    :  4    :  :  20  :  10) 

I  6.  by  multiplication,  rdirb:  :e:d (2r  :  6r  :  :    5:15) 

I  r.  by  division,  ^     ±  :  1.  :  :  e  :  d  (— :  —  :  i    5:15) 
V.  r      r  r       r 

Because  the  product  of  the  means,  in  each  case,  is  equal  to 

that  of  the  extremes,  and  therefore  the  quantities  ai^  pro-^ 

portional,  by  theorem  2, 

L 


74  Of  Proportian. 

THEOREM  VL 

If  three  numbers^  r/,  h^  c  (2, 4,  8),  be  in  continued propor- 
tionj  the  square  of  the  first  xvill  be  to  that  of  the  second  as 
the  first  number  to  the  third ;  that  is^  c^  \  b^  \  \  a  ;  c. 

For,  since  a  \  b  \  \  b  \  c^  thence  will  ac  =  bby  by  theorem 
1  ;  and  therefore  aac  ==  abb^  by  equal  multiplication  ;  con- 
sequently a^  :  b^  :  :  a  :  Cy  by  theorem  2. 

In  like  manner  it  may  be  proved,  that  of  four  quantities 
continually  proportional,  the  cube  of  the  first  is  to  that  of 
the  second  as  the  first  quantity  to  the  fourth. 

THEOREM  VII. 

In  any  continued  geometrical  proportion  (1,  3,  9,  27,  81^ 
<if  e.)  the  product  of  the  two  extremes^  and  that  of  every 
oHier  tzvo  terms^  equally  distant  fro^n  them^  are  equaU 

For  the  ratio  of  the  first  term  to  the  second,  being  the 
same  as  that  of  the  last  but  one  to  the  last,  these  four  term^ 
are  in  proportion  ;  and  therefore,  by  theorem  1 ,  the  rect- 
angle of  the  extremes  is  equal  to  that  of  their  two  adjacent 
terms  ;  and,  after  the  very  same  manner,  it  will  appear, 
that  the  rectangle  of  the  third  and  last  but  two  is  equal  to 
that  of  their4;wo  adjacent  terms,  the  second  and  last  but 
one  ;  >  and  "so  of  the  rest.  Whence  the  truth  of  the  propo- 
sition is  mnnitest. 

THEOREM  VIII. 

The  sum  of  any  number  of  quantities^  in  continued  geome- 
trical proportion^  is  equal  to  the  difference  of  the  rectangle 
of  the  second  and  last  terms  ^  and  the  square  of  the  first  ^  di- 
vided by  the  difference  of  the  first  and  second  terms. 

For,  let  the  first  term  of  the  proportion  be  denoted 
by  rt,  the  common  ratio  by  r,  the  number  of  terms  by 
w,  and  the  sum  of  the  whole  progression  by  x\  then  it 
is  manifest  that  the  second  term  will  be  expressed  by  a  X  r^ 
or  ar  ;  the  third  by  ar  x  r,  or  ar'^  ;  the  fourth  by  ar"^  X  ^, 
or  «r^,  and  the  ;2th  or  last  term  by  ar^"'^ ;  and  there- 
fore the  proportion  will  stand  thus  :  a  +  ar  +  ar^  -f  ar^ 
.    .    .    •  -f  ar'^"^  +  ar^"'^  =  x ;    which  equation,  mul- 


Of  Proportion.  75 

tiplied  by  r,  gives  ar  +  ar^  -f  <2r^  +  ar"^  ......+  «r^^-^ 

4-  ar^  =  rx ;    from  which  the  first  equation  being  sub- 
tracted there  will  remain  — a  +  ar'^  =^  rx  ---^  x  ;  whence 

/ar^  —  a      r  X  ar^"^  —  a\       ar  X  ar^^"^  —  aa 

X  =  { = I  == ;    as 

\  r  —  1  r  —  1         /  ar  —  a 

was  to  be  demonstrated. 


SECTION  XL 


The  Application  of  Algebra  to  the  Resolutioji  of 
Numerical  Problems, 

WHEN  a  problem  is  proposed  to  be  solved  algebrai- 
cally, its  true  design  and  signification  ought,  in  the  first 
place,  to  be  perfectly  understood,  so  that,  if  needful,  it  may 
be  abstracted  from  all  ambiguous  and  unnecessary  phrases, 
and  the  conditions  thereof  exhibited  in  the  clearest  light 
possible.  This  being  done,  and  the  several  quantities 
therein  concerned  being  denoted  by  proper  symbols,  let 
the  true  sense  and  meaning  of  the  question  be  ti'anslated 
from  the  verbal  to  a  symbolical  form  of  expression;  and 
the  conditions,  thus  expressed  in  algebraic  terais,  will,  if 
it  be  properly  limited,  give  as  many  equations  as  are  ne- 
cessary to  its  solution.  But,  if  such  equations  cannot  be 
derived  without  some  previous  operations,  which  fre- 
quently happens  to  be  the  case,  then  let  the  learner  ob- 
serve this  rule,  viz,  let  him  consider  what  method  or  pro- 
cess he  would  use  to  prove,  or  satisfy  himself  in,  the  truth 
of  the  solution,  were  the  numbers  that  answer  the  condi- 
tions of  the  question  to  be  given  or  affirmed  to  be  so  and 
so  ;  and  then,  by  following  the  very  same  steps,  only  using 
unknown  symbols  instead  of  kilown  numbers,  the  question 
will  be  brought  to  an  equation. 

Thus,  if  the  question  were  to  find  a  number,  which 
being  multiplied  by  5,  and  8  subtracted  from  the  pro- 


ax  —  h 
2axb  +  bK 


7b  The  Application  of  Algebra 

duct,  the  square  of  the  remainder  shall  be  144  ;  then,  hav- 
ing put  a  =  5,  ^  =  8,  and  c  =  144,  suppose  the  niimber 
sought 

to  be         -  -  -  4    (or) 

then  5,  or  a  times  that  number  1        ^o 

will  be  J 

from  which  8,  or  b^  being  sub- 1         . 

tracted,  there  remains  J 

which  squared  is         -  -         144 

Therefore  a^x^  —  2axb  +  b^  is  =z  c  (or  144),  according 
to  the  conditions  of  the  question.  In  the  same  manner 
may  a  question  be  brought  to  an  equation  when  two  or 
more  quantities  are  required. 

After  the  conditions  of  a  problem  are  noted  down  in 
algebraic  terms,  the  next  thing  to  be  done  is  to  consider 
whether  it  be  properly  limited,  or  admits  of  an  indefinite 
number  of  answers  ;  in  order  to  discover  which,  observe 
the  following  rules, 

RULE  r. 

When  the  nu7nber  of  quantities  sought  exceeds  the  num- 
ber of  equations  given^  the  question^  for  the  ^ general  party 
7S  capable  of  innumerable  answers. 

Thus,  if  it  be  required  to  find  two  numbers  (x  and  y) 
with  this  one  single  condition,  that  their  sum  shall  be  100, 
we  shall  have  only  one  equation,  viz.  x  +y=  100,  but  two 
unknown  quantities,  x  and  2/,  to  be  determined  ;  therefore 
it  may  be  concluded  that  the  question  will  admit  of  innu- 
merable answers, 

RULE  IL 

But  if  the  number  of  equations^  given  from  the  conditions 
of  the  question^  be  just  the  same  as  the  number  of  quantities 
s ought y  then  js  the  question  truly  limited. 

As,  if  the  question  were  to  find  two  numbers,  whose 
sum  is  100,  and  whose  difference  is  20 ;  then,  x  being 
put  for  the  greater  n^timber,  and  y  for  the  less,  we  shall 
have  a;  -f  z/  =  100,  and  x  —  z/  =  20:  therefore,  there 
being  here  two  equations  and  two  unknown  quantities, 


to  the  Re&olutnon  of  Problems.  77 

the  question  is  truly  limited ;    60  and  40  being  the  only 
two  numbers  that  can  answer  the  conditions  thereof. 

RULE  III. 

When  the  number  of  equations  exceeds  the  number  of 
quantities  sou^ht^  either  the  conditions  of  the  problem  are 
inconsistent  one  with  another^  or  what  is  proposed  iii  gene-- 
ral  terms  can  only  be  possible  in  certain  particular  cases. 

But  it  is  to  be  observed,  that  the  equations  understood 
here,  as  well  as  in  the  preceding  rules,  are  supposed  to 
he  no  ways  dependent  upon,  or  consequences  of  one  ano- 
ther. If  this  be  not  the  case,  the  question  may  be  either 
unlimited  or  absurd,  or  perhaps  both,  at  the  same  time 
that  it  seems  truly  limited  \  as  will  appear  by  the  follow- 
ing example : 

Wherein  it  is  required  to  find  three  numbers,  under 
these  conditions,  that  the  sum  of  once  the  first,  twice 
the  second,  and  three  times  the  third,  may  be  equal  to  a 
given  number  b  ;  that  the  sum  of  four  times  the  first, 
five  times  the  second,  and  six  times  the  third,  may  be 
equal  to  a  given  number  c ;  and  that  the  sum  of  seven 
times  the  first,  eight  times  the  second,  and  nine  times 
the  third,  may  be  equal  to  a  third  given  number  d.  Now, 
the  three  numbers  sought  being  respectively  denoted  by 
X,  y,  and  2,  the  question,  in  algebraic  terms,  will  stand 
thus : 

;^^  +  2z/  -f-  32  =  Z» 
4:?^  +  5z/  -f-  62  =  f 
7x  +  Sy  +  9z  =  d. 
Here,  there  being  three  equations,    and  just  the  same 
number  of  unknown  quantities,  one  might  conclude  the 
question  to  be  truly  limited:    but,  by  reflecting  a  little 
upon  the  nature  and  form  of  these  equations,  the  con- 
trary will  soon  appear  ;  because  the  last  of  them  includes 
no  new  condition  but  what  is  comprised  in  and  may  be 
derived  from  the  other  two  ;    for  if  from  the  double  of 
the  second  the  first  equation  be  taken  away,  the  value 
of  7x  +  8y  -f  92  will  from  thence  be  given  =  2c  -^  ^,  ■ 
Hence  it  is  manifest,  that  giving  the  value  of  7x  +  Si/ 
+  92,    in    the   third  equation,    contributes   noticing  to- 


7^  The  Application  tf  Algebra 

wards  limiting  the  problem ;  and  that  the  problem  itself 
is  not  only  unlimited,  but  also  impossible,  except  when  d 
is  given  equal  to  2c  -—  b. 

Having  laid  down  the  necessary  rules  for  bringing  pro- 
blems to  equations,  and  for  discovering  when  they  are 
truly  limited,  it  remains  that  we  illustrate  what  is  hither- 
to delivered  by  proper  examples. 

Arithmetical  Problems. 

PROBLEM  I. 

To  find  that  number^  to  which  7  S  being  added^  the  sum 
.shall  be  the  quadruple  of  the  said  required  number* 

Let  the  number  sought  be  represented  by  -  a" ; 

then  will  its  quadruple  be  denoted  by  -  Ax  ; 

whence,  by  the  conditions  of  the  question,      a;  +  75  =  4^^ ; 

this  equation,  by  transposing  a:,  becomes  75  =  3x  : 

75 
from  whence,  dividing  by  3,  we  have  at  =  —  =25, 

o 

which  is  the  number  that  was  to  be  found  j  for  it  is  plain 
that  25  +  25  X  3  =  25  X  4  =  100. 

PROBLEM  IL 

What  number  is  thatj  which  being  added  to  4,  and  also 
multiplied  by  4,  the  prb duct  shall  be  the  triple  of  the  sum  7 

Let  the  number  sought  be  denoted  by  -  x\ 

so  shall  the  sum  be  denoted  by  -  a:  +  4, 

and  the  product  by  -  -  -  Ax : 

whence,  by  the  conditions  of  the  question,  4x  =z  x  -^  4 
X  3  ;  that  is,  4x  z=z  3x  +  12  ;  from  which,  by  transposi- 
tion, a;  =  12. 

PROBLEM  in. 

To  find  txvo  numbers  such^  that  their  sum  shall  be  30,  and 
their  difference  12. 

If  X  be  taken  to  denote  the  lesser  of  the  two  numbers  ; 
then,  by  adding  the  difference  12,  the  greater  number  will 
be  denoted  by  -%•  -f  12  ;  and  so  we  shall  have  2Ar  +  12  = 
30,  by  the  question. 

From  which  equation,  2.r  =  30  —  12  =  18;  and  con- 


io  the  Resolution  of  Problems,  79 

18 
sequently,  ::c  =  —  =  9 ;     whence  the  greater  number 

A* 

{x  +  12)  is  also  given  =  21. 

PROBLEM  IV. 

To  divide  the  number  60  into  three  such  parts^  that  the 
first  may  exceed  the  second  by  8,  and  the  third  by  16. 

Let  the  first  part  be  denoted  by  x  ;  then  the  second  will 

be  X  —  8,  and  the  third  x  —  16:    the  aggregate  of  all 

which,  or  ^x  —  24  is  r=  60,  by  the  question* 

84 
Hence  3x  =  60  +  24  =  84,  and  ^  =  —  =  28  :   so  that 

3 

28,  20,  and  12  are  the  three  parts  required. 
PROBLEM  V. 

The  sum  of  660/.  xvas  raised^  for  a  certain  purpose^  by 
four  persons^  A,  B,  C,  and  D  :  whereof  B  advanced  twice 
as  much  as  A  i  C  as  much  as  A  and  B  ;  a?id  D  as  much 
as  B  and  C  :  what  did eachpersoyi  contribute? 

Let  the  sum  or  number  of  pounds  advanced  by  1 

A  be  called  '  J  ^ ' 

then  will  the  number  of  B's  pounds  be  denoted  by         ^x  ; 
that  of  C's  by  -  -  -  Zx  \ 

and  that  of  D's  by  -  -  5x  y 

the  sum  of  all  which  is  given  equal  to  660/.  that  is^  \\x 

=  660 :  from  whence  x  = =  60.     Therefore  60,  120, 

11  '        ' 

180,  and  300/.  are  the  respective  sums  that  were  to  be  de- 

termined,  ^* 

PROBLEM  VL 

A  certain  simi  of  money  was  shared  among  five  persons^ 
A,  B,  C,  D,  and  E  ;  whereof  B  received  10/.  less  than  A  ; 
C  16/.  more  than  B  ;  D  5/.  less  than  C  ;  E  15/.  more  than 
D  :  moreover  it  appeared  that  the  shares  of  the  txvo  last 
together  xvere  equcd  to  the  sum  of  the  shares  of  the  other 
three :  what  xvas  the  whole  sw7i  shared^  and  how  mxich  did 
each  receive  ?  , 


>will  be  the  share  of  <^  j^ 

I      ' 
LE; 


80  77ie  Application  of  Algebra 

Let  X  denote  the  share  of  A  : 

then^'"+    f 

and  therefore  2x  -f-17  =  3x  —  4,  ^z/  the  question:  from 
whence,  by  transposition,  21  =  at  ;  so  that  21,  11,  27,  22, 
and  37/.  are  the  several  required  shares  j  amounting,  in  the 
whole,  to  118/. 

PROBLEM  VIL 

To  find  three  numbers  on  these  conditions^  that  the  stint 
of  the  first  and  second  shall  be  15  ;  of  the  first  and  third 
16  ;  and  of  the  second  and  third  1 7. 

If  the  first  number  be  denoted  by  a;  ;  then  it  is  plain; 
by  the  question,  that  the  second  will  be  represented  by 
15  — x^  and  the  third  by  16  —  x.  But  the  sum  of  these 
two  last  is  given  equal  to  17;  that  is,  31  — 2.v  =17; 
whence,  by  transposition,  14  =  2^; ;  and  consequently  x  = 

14 

—  =  7.  Hence  15  —  :v  =  8,  and  16  —  .v  =  9 ;  which  are 
2 

the  other  two  numbers  required. 

PROBLEM  VIIL 

To  find  that  number y  -which  being  doubled^  and  16  sub- 
tracted from  the  product^  the  remainder  shall  as  much  ex- 
ceed 100  as  the  required  number  itself  is  less  than  100. 

The  number  sought  being  denoted  by  x^  the  double 
thereof  will  b/x  represented  by  2x ;  from  which  subtracts- 
ing  16,  the  remainder  will  be  2x —  16;  and  its  excess 
above  100  equal  to  2x  —  16  —  100 :    therefore  2x  — ?  16 

—  too  =  100  —  X,  bij  the  question;  whence  3.v  =  216  ; 

1  216 

and  consequently  x  =  -^  =  72. 
3 

PROBLEM  IX. 

7o  divide  the  number  75  into  two  such  parts  that  three 
times  the  greater  may  exceed  seven  times  the  lesser  by  15. 
Let  the  greater  part  be  =  a:  ;    then  will  the  lesser 


io  the  Resolution  of  Problems.  81 

part  =  75  — .r,  and  we  shall  have  Sx  — 15  =  7S*-^x 
X  T ;  or,  which  is  the  same  thing,  Sx  —  15  =  525  —  7x  : 
from  whence  10.v  =  540,  and  conseqilently  :v  =  54. 

PROBLEM  X. 

Two  persons^  A  and  B,  having  received  equal  sums  of 
money ^  A  out  of  his  paid.away  251.  aiid  B  of  his  60/.  and 
then  it  appeared  that  A  had  just  truice  as  much  money  as  B  : 
what  money  did  each  receive  ? 

Suppose  .V  to  denote  the  sum  received  by  each  per- 
son ;  then  A,  after  paying  away  25/.  had  x  —  25  ;  and  B, 
after  paying  away  60/.  had  x  —  60  :   hence  x  —  25  =  2x 

—  120,  by  the  question;  and  therefore  120  —  25  =  2x  —  .v^ 
that  is,  95  =  X. 

PROBLEM  XL 

To  find  that  number  whose  \  part  exceeds  its  \  part  by 
12. 

Let  the  number  sought  be  represented  by  x ;  then  will 

XX 

=  12,  by  the  conditions  of  the  problem  ;  which 

equation,  by  multiplying  every  numerator  into  all  the  de- 
nominators except  its  o\vn,  gives  Ax  —  3^  =  144,  that  is, 
X  =  144. 

PROBLEM  XIL 

What  sum  of  motley  is  that  xuhose  ^  party  i  fia^'t^  and  -|- 
party  added  together ^  shall  amount  to  94  pounds  ? 

If  X  be  the   number  of  pounds  required,    then  will 

v  X  X 

1 — I f-  ^ —  =  94 :   from  whence,  by  reduction,  20;\:  -f 

15^  +  12;f  =  94  X  60,  that  is,  ATx  =  94  X  60  ;  and  there.- 
fore..v  =  2  X  60=  120. 

PROBLEM  XIIL 

In  a  mixture  of  copper  j  tin^  and  lead ^  one  half  of  the  whole 

—  16lb.  was  copper  ;  one  third  of  the  whole  —  12lb.  tin  ; 
and  one  fourth  of  the  xvhole  +  4lb.  lead :  -what  quantity  of 
each  7vas  there  in  the  composition  P 

'  M 


82  7  lie  Application  of  Algebra 

Litt  oc  denote  the  weight  of  the  whole  : 


then  will 


V       I  i 

<|  —  —  12  ^bethe  weight  of  the  <i 
3  I 


tm, 

-+    4 
L  4^  J  Uead ; 

iiud,    if   all   these    be    added    together,    we    shall   have 

V  ')C  SC 

1 — I 1 —  24  =  Xj  by  the  questioii.      Hence,  by 

reduction,  12x  +  8^  +  6a:  —  576  =  24;v ;  therefore 
2a;  =  576,  and  a;  ==  —  =  288.  So  that  there  were  128lb. 
of  copper,  84lb.  of  tin,  and  76lb.  of  lead. 

PROBLEM  XIV. 

What  Slim  of  money  is  that^  from  which  51.  being  suh<= 
tract  ed^  txvo-thirds  of  the  remainder  shall  be  40/.  ? 

Let  X  represent  the  required  sum  ;  then,  5  being  sub- 
tracted, there  will  remain  x  —  S  \    two-thirds  of  which 

■  ,  I                        giX"  —  10 
will  be  a:  —  5  x  f ,  or ;  and  so,  by  the  question^ 

'^x  -—  10 

we  have =  40:   whence  %x — 10  =  120;  and 

3 

-- —  =  65 


PROBLEM  XV. 

What  number  is  that^  rvhich  being  divided  by  12,  the 
quotient^  dividend^  and  divisor^  added  all  together^  shall 
amount  to  M^ 

Let  X  =  the  required  number  ;  so  shall 

f-  :)^  -}-  12  =  64,  by  the  conditions  of  the  question. 

Whence  x  +  \2x  =  52  x  12,  or  13;c  =  624 ;  and  conse- 

1  624       .„ 

quently  x  =:  — ^  =  48. 


tsi  the  Resolution  of  Problems.  83 


PROBLEM  XVI. 

Tofmdttvo  numbers  in  the  proportion  of  2  to  t^  so  thaty 
//4  be  added  to  each^  the  two  sums  thence  arising'  shall  be 
in  proportion  as  3  to  2. 

Let  X  denote  the  lesser  number ;  then  the  greater  will 
be  denoted  by  2x  ;  and  so,  by  the  question^  we  shall  have 
2x  +  4<  ;  X  +  4^  :  :  3  :  2.  From  whence,  as  the  product 
of  the  two  extremes  of  any  four  proportional  numbers  is 
equal  to  the  product  of  the  two  means  (see  section  10,  the- 
orem 1),  we  have  the  following  equation,  viz.  2x  +  4X2 
=  a'  -f  4  X  3,  that  is,  4Ar  +  8  =  3Ar  +  12  :  whence  x  =i  4^^ 
and  2x  =  8  :  which  are  the  two  numbers  that  were  to  he 
found. 

PROBLEM  XVIL 

A  prize  of  20001.  was  divided  between  two  persons ^  whose 
shares  therein  were  in  proportion  as  7  to  9  :  xvhat  was  the 
share  of  each  ? 

If  ^  =  the  share  of  the  first,  then  that  of  the  second  will 
be  2000  —  X  ;  and  we  shall  have  x  :  2000  —  x\\7\9. 

Hence,  by  multiplying  the  extremes  and  means,  9x  = 

14000 — 7x\  from  which  x  is  found  =  =  875/. 

16 

and  2000  —  AT  =  1 125/. 

PROBLEM  XVIIL 

A  bill  of  1201.  was  paid  in  guineas  and  moidores^  and  the 
number  of  pieces  of  both  sorts  was  just  100  ;  to  find  hozv 
many  there  were  of  each. 

If  X  =  the  number  of  guineas,  then  will  100  —  x  be 
the  number  of  moidores  :  therefore  the  number  of  shil- 
lings  in  the  guineas  being  21;^',  and,  in  the  moidores,  27 
X  100  ~  X,  we  have  21;^'  +  27  X  100  —  x  =  120 
X"  20  =  the  shillings  in  the  whole  sum :  hence,  by 
multiplication,  2\x  +  2700  —  27.v  =  2400;  and  x  = 
300 
-  =  50. 


84  The  Application  of  Algebra 

PROBLEM  XIX. 

A  labourer  engaged  to  serve  40  days^  on  these  conditions^ 
that  for  every  day  he  xvorked  he  rvas  to  receive  20  pence  ^  but 
that  for  every  day  he  played^  or  xvas  absent^  he  zvas  to  forfeit 
8  pence  ;  noxv^  after  the  40  days  xvere  expired^  it  xvas  found 
that  he  had  to  receive^  upon  the  xvhoky  380 pence:  the  ques- 
tion isy  tofindhoxv  many  days  of  the  forty  he  xvorked^  and 
hoxv  many  he  played. 

Let  the  number  expressing  the  days  he  worked  be  re- 
presented by  :c  ;  then  the  number  of  days  he  played  will 
be  expressed  by  40  —  x\  moreover,  since  he  was  to  re- 
ceive 20  pence  for  every  day  he  worked,  the  whole  num- 
ber of  pence  gained  by  working  will  be  20a;  ;  and,  for 
the  like  reason,  the  number  of  pence  forfeited  by  play- 
ing, or  being  absent,  will  be  8  X  40  —  at,  or  320  —  %x  \ 
which  deducted  from  20^^,  leaves  28.v  —  320,  for  the  sum 
total  of  what  he  had  to  receive :  whence  we  have  this 
equation,  28Ar  —  320  =  380 :   from  which  28;c  =  380  -f- 

320  =  700,  and  consequently  x  =  —  =  25,  equal  to  the 

28 

number  of  days  he  worked  ;   therefore  40  —  25  =  15  will 

be  the  number  of  days  he  played. 

PROBLEM  XX. 

A  farmer  xvould  mix  txvo  sorts  of  grain ^  vr/..  -ivhcai. 
xvorth  45.  a  bushel^  xvith  rye^  xvorth  2s.  6d,  the  bushel^  so 
that  the  xvhole  mixture  may  consist  oj  100  bushels^  and  be 
xvorth  3^.  2d.  the  bushel:  noxv  it  is  required  to  find  hoxv 
many  bushels  of  each  sort  must  be  taken  to  make  up  such  a- 
mixture. 

Let  the  number  of  bushels  of  wheat  be  put  =  .v,  and 
the  number  of  bushels  of  rye  will  be  100  —  x  :  but  the 
number  of  bushels  multiplied  by  the  number  of  pence 
per  bushel,  is  equal  to  the  number  of  pence  the  whole 
is  worth ;  therefore  48a:  is  the  whole  value  of  the  wheat, 
and  30  X  100  —  ~,  or  3000  —  30.x,  that  of  the  rye  ; 
and,  consequently,  48.v  -f-  3000  —  30a,^,  the  sum  of  these 
two,  the  whole  value  of  the  mixture :  which,  by  the 
question^  is  equal  to  lOO  X  38,  or  3800  pence:    hence 


to  the  Resolution  of  Problems.  85 

we  have  48;^  +  3000  —  30.v  =  3^00 ;    and  therefore 

800 
=   —  =  44^,    the    number   of  bushels  of   wheat; 
18  ^' 

whence  the  number  of  bushels  of  rye  will  be  100  —  44| 
=  55f- 

PROBLEM  XXL 

A  farmer  sold  to  one  7nan  30  Imshels  of  xvheat  and  40 
of  barley^  and  for  the  whole  received  270  shillings  ;  and  to 
another  he  sold  50  bushels  of  wheat  and  30  of  barley^  at  the 
same  prices^  and  for  the  whole  received  340  shillings :  now 
it  is  required  to  find  what  each  sort  of  grain  was  sold  at  per 
busheL 

Let  X  and  y  be  respectively  the  number  of  shillings 
which  a  bushel  of  each  sort  was  sold  for  ;  then,  from  the 
conditions  of  the  question,  we  shall  have  these  two  equa- 
tions, viz. 

SOx  +  40y  =  2rO, 
5Qx  +  SOy  =  340  ; 
from  four  times  the  second  of  which  subtrajct  three  times 

550 
the  first,  so  shall  llO.r  =  550  ;  and  consequently  x  =  

=  5  ;   moreover,  by  subtracting  3  times  the  second,  from  . 
5  times  the  first,  you  will  have  110;/  =  330,  and  therefore 

^  ""  no"" 

For    /  ^0  X  5  +  40  X  3  =  270, 
150X5  +  30X3  =  340. 

PROBLEM  XXIL 

A  son  asking  his  father  how  old  he  xvas^  received  the 
folloxving  reply  :  My  age^  says  the  father^  7  years  ago^  xvas 
just  four  times  as  great  as  yours  at  that  time  ;  but^  7  years 
hence^  if  you  and  I  live  ^  my  age  will  then  be  only  double  of 
yours:  it  is  required  to  find  from  hence  the  age  of  each 
person. 

Let  X  represent  the  age  of  the  son  seven  years  be- 
fore the  question ;  then  the  age  of  the  father,  at  that 
time,  was  4.r,  by  the  conditions  of  the  question  j  and,  if 


8  5  i  'he  Applicatmn  of  Algebra 

each  of  these  ages  be  increased  by  14,  it  is  plain  that 
<%•  +  14  and  4iX  +  14  will  respectively  express  the  two 
ages  7  years  after  the  time  in  question ;  whence,  again, 
by  the  problem,  we  have  4^x  +  14  =  2  X  iv  -f  14  ;  from 
which  ^  =  7,  and  4x  =  28 ;  therefore  7  +  7  =  14,  and 
28  +  <^  =  35,  are  the  two  ages  required. 


For   / 35  — 7=14-7x4, 
I  35  4*  7  =  14  4-  7X2. 


35  4*  7  =  14  -f  7X2. 

PROBLEM  XXIIL 

A  gentleman  hired  a  servant  for  12  months^  and  agreed 
to  allow  him  20/.  and  a  livery^  if  he  staid  till  the  year  was 
expired;  but  at  the  end  of  8  months  the  servant  -went 
away^  and  received  12 L  and  the  livery^  as  a  proportional 
part  of  his  wages :  the  question  is^  what  was  the  livery 
valued  at  ? 

Let  X  be  the  value  sought ;  then  20  +  :v  will  be  the 
whole  wages  for  12  months,  and  12  +  at  the  part  thereof 
which  he  received  for  8  months. 

But  the  wages  being  in  the  same  proportion  as  the 
times  in  which  they  are  eai'ned,  or  become  due,  we  there- 
fore  have,  as  12  :  8  :  :  20  +  .y  :  12  +  ;c ;  whence  12  X 
12  +  ^  !=  8  X  20  4-  x^  or  144  +  12.v  =  160  +  %x  (by 
tli^or.  1,  p.  72),  consequentlv  12.v  —  8v  =  160—144, 

and  .V  =  —  =  4/. 

4 

PROBLEM  XXIV. 

Fc'ur  persons^  A,  B,  C,  D,  spent  twenty  shillings  in  com.- 
pany  together  ;  vohereof  A  proposed  to  pay  J,  B  -J-,  C  |, 
G7id  D  ^  part ;  Imt^  when  the  money  came  to  be  collected^ 
they  found  it  was  not  sufficient  to  answer  the  intended  pur- 
pose :  the  question  then  is^  to  find  liQw  much  each  person 
7niLst  contribute^  to  viake  up  the  whole  reckonings  supposing 
their  several  shares  to  be  to  each  other  in  the  proportion 
above  specified. 

Let    X    be    the    share  of    A ;      then    it   will    be,    as 

}  :  -J^  or  4  :  3  :  :  \'  :  —  zz:  the  siiarc  of  B  ;    and,  as 


to  the  Resolution  of  Problems.  ST 

\  :  I,  or,  as  5  :  3  :  :  .v  :  ~  =  the  share  of  C  ;    also,  as 

^  :  I,  or,  as  2  :  1  :  :  AT :  —  =  the  share  of  D.  ' 

Therefore,  by  the  question^  x  •\ 1 f--^  =  20; 

whence  mx  +  ZOx  +  24a;  +  20:^;  =  800,  that  is,  114.r 

=  800 ;    and  consequently  x  =  - —  =  7-5V?  the  share  of 

3iV 
A  J  therefore  (^)  that  of  B  will  be  =  5|-f :  that  of  C 

(|:)  =:  4lf  ;  and  that  of  D  (-|)  .=  3|f 

PROBLEM  XXV. 

A  mar ket'Xv Oman  purchased  a  certain  number  of  eggs  at 
2  a-penny^  and  as  many  at  3  a-penny^  and  sold  them  all  out 
again  at  the  rate  of  5  for  tixn  pence ^  and  lostfohr  pence  by 
so  doing :  what  number  of  eggs  did  she  buy  and  sell? 

Let  X  be  the  number  of  eggs  of  each  price,  or  sort ; 

X 

then  —  will  be  the  number  of  pence  which  all  ,the  "first 

X 

sort  cost,  and  —  the  pric^  of  all  the  second  sort ;    but 

the  whole  price  of  both  sorts  together,  at  the  rate  of 

5    for    two  pence,    at    which    they  were  sold,    will  be 

^x  4  V 

— ,  for  as  5  :  2  :  :  2x  (the  whole  number  of  eggs)  :.  —  • 

hence,  by  the  question^  ^ — |- =  4  ;  whence  1 5.v 

+  \Qx  —  24;c  =  120,  and  therefore  xr=i\2Q. 

Forl22  +  l!2._!12x2  =  60+40-96^4. 
2^35 

PROBLEM  XXVI. 

A  composition  of  copper  and  tin^  containing  100  cubic 
ifichesy  being  weighed^  its  xveight  -wasfoxmd  to  be  505  oiincef^ : 


88  The  Application  oj  Algebra 

how  many  ounces  of  each  metal  did  it  contahi^  supposing  the 
weight  of  a  cubic  inch  of  copper  to  be  5^  ounces^  and  that  of 
a  cubic  inch  of  tin  4t\? 

Let  X  be  the  number  of  ounces  of  copper ;  then 
505  —  X  will  be  the  number  of  ounces  of  tin,  and  we  shall 
have 

5\\  1  (cubic  inch)  -,  :  x  i  ^  inches  of  copper, 

4  J  :  1  (cubic  inch)  :  :  505  —  x  : lH-  inches  of  tin, 

X  505  X 

Therefore,  ~   -| =  100,  by  the  question. 


Whence  4^  x  a,^  +  5|  X  505  —  x  =  5i  X  4i  X  100,  that  is, 
17X  V      21X505— A7        21  X  irx  100^  _  21  X  17X25 

4  4  ^""  4x4  ^'^  4  ' 

which,  by  rejecting  the  common  divisor,  becomes   17x 
+  21  X  505— "Iv  =  21  X  ir  X  25  =  8925,  or  17x  —  21.v 

=  8925  -^  10605  =  —  1680.     From  whence  x  =  ^^ 

4 

z=  420 ;  and  505  —  x  z=:  85  -,  whicli  are  the  two  numbers 
required. 

The  same  otherxvise. 

Suppose  .V  to  be  the  number  of  solid  inches  of  cop- 
per ;  the;n  the  number  of  inches  of  tin  being  100  —  x^ 
we  have  5^  X  ^  +  4^1  X  100  —  x  =  505,  that  is, 
■51 X  +  425  —  41;c  =  505,  or  x  =z  505  —  425  =  80; 
v/hich,  multiplied  by  5^,  gives  420,  for  the  ounces  of 
copper. 

PROBLEM  XXVIL 

A  shepherd^  in  time  of  zuar^  fell  in  ivith  a  party  cf  sol- 
diers^ who  plundered  him  of  half  liisflock^  and  half  a  sheep 
over ;  afterwards  a  second  party  met  him^  who  took  half 
xvhat  he  had  left^  and  half  a  sheep  over ;  and^  soon  after 
this^  a  third  party  niet  Imn^  and  used  him  in  the  same  7nan- 
7ier;  and  then  he  had  only  five  sheep  left :  it  is  required  to 
find  xvhat  number  of  sheep  he  had  at  firsts 


to  the  Resolution  of  Problems,'  89 

htt  X  (as  usual)  be  the  number  sought ;  then,  ac- 
cording to  the  question,  the  number  of ^  sheep  left,  after 
being  plundered   the    first  time,    will   be    expressed  by 

1    or  •-'        ■  ;    the  half  of  which  is  — ^^^^^  ;    from 

X     .1  1             X  — —  3 
whence  subtracting  |,  the  remainder,  (-^ —  |) ^ 

will    be    the  number  of   sheep   left    after    being    plun- 

;!^   3 

dered  the  second  time  :    in  like  manner,  if  from 

'  8 

^  3 

(the  half  of )  you  again  take   i,  there  will  re- 

X 3  X  — —  7 

main ^,  or  — —  ,  the  number  of  sheep  remain- 

8  8 

X 7 

ing  at  last.     Hence  we  have =  5  ;  therefore  x  —  7 

8 

=  40,  and  .%•  =  47. 

PROBLEM  XXVIIL 

The  difference  of  txvo  numbers  being  given .^  equal  to  4, 
and  the  difference  of  their  squares^  equal  to  40  ;  to  find  the 
numbers. 

Let  the  lesser  number  be  x ;  then,  the  difference  be- 
ing 4,  the  greater  must  consequently  be  x  -f  4,  and  its 
square  xx  +  8^  +  16,  from  which  xx^  the  square  of  the 
lesser  being  taken  away,  the  difference  is  8j^  +  16  :  there- 
fore 8x  +  16  =  40  ;  which,  reduced,  gives  at  =  3  j  w^hence 
.T  -f-  4  =  7 ;  therefore  the  two  required  numbers  are  3 
and  7. 

All  the  problems  hitherto  delivered  are  resolv^ed  by  a 
numeral  exegesis^  wherein  the  unknown  quantities  only 
are  represented  by  letters  of  the  alphabet ;  which  seem- 
ed necessary,  in  order  to  strengthen  the  beginner's 
idea^  at  setting  out,  and  lead  him  on  by  proper  gra- 
dations :  but  it  is  not  only  more  masterly  and  elegant^ 
but  also  more  useful,  to  represent  the  known,  as  well 
as  the  vmkuown  quantities,  by  algebraic  symbols ;    since 

N 


jg  ihe  Application  of  Algebra 

from   thence    a    general    theorem    is    derived,    whereby 
all  other  questions  of  the  same  kind  may  be  resolved. 

As  an  instance  hereof,  let  the  last  problem  be  again 
resumed ;  then  the  given  difference  of  the  required 
numbers  being  denoted  by  a,  the  difference  of  their 
squares  by  b^  and  the  lesser  number  by  .v ;  the  greater 
will  be  a:  -f-  a,  and  its  square  x^  +  2;^a  -f  d^ ;  from 
which  a'2,  the  square  of  the  lesser  number,  being  de- 
ducted, there  remains  2xa  -^  a^  z=  b :  whence,  if  aa 
be  subtracted  from  both  sides,  there  will  remain  2ax  = 

b  —  aa;  this,  divided  by  2«,  gives  x  ^= ;    and 

consequently,  .v  -f  ^  =  —  -| .       Hence  it  appears, 

that,  if  the  difference  of  the  squares  be  divided  b}- 
twice  the  difference  of  the  numbers,  and  half  the  dif- 
ference of  the  numbers  be  subtracted  from  the  quotient, 
the  remainder  will  be  the  lesser  number ;  but  if  half 
the  difference  of  the  numbers  be  added  to  the  quotient, 
the  sum  w^ll  give  the  greater  number.  Thus,  if  the  dif- 
ference (a)  be  4,  and  the  difference  (^)  of  the  squares 

40  (as  in  the  case  above)  ;     then    ( — )  the    difference 

of  the  squares,  divided  by  twice  the  difference  of  the 
numbers,  will  be  5  ;  from  which  subtracting  (2)  half 
the  difference  of  the  numbers,  there  remains  3,  for  the 
lesser  number  sought ;  and  by  adding  the  said  half  dif- 
ference, you  will  have  7  =  the  greater  number.  In  the 
same  manner,  if  the  difference  of  the  two  numbers  had 
been  given  6,  and  tiie  difference  of  their  squares  60,  the 
numbers  themselves  would  have  come  out  2  and  8  :  and 
so  of  any  other. 

PROBLEM  XXIX. 

Havhi^  given  the  su?7i  of  tzvo  7iitmbcrs^  equal  to  30,  and 
the  difference  of  their  squares^  equal  to  120  ;  to  find  the 
numbers. 

Put  a  =  30,  ^  =  120,  and  let  x  be  the  lesser  num- 
ber sought,   then  the  greater  will   be    a  —  x  \    whose 


to  the  Resolution  of  Frohkms.  91 

square  is  aa  —  2ax  +  o(^  ;  from  which  the  square  of  the 
lesser  being  subtracted,  we  have  a^  —  2ax  =  b  ;  this  re- 

(iiced  c-ives  .v,  the  lesser  number,  =  ~ =  13. 

Therefore  the  greater  {a  —  ;:)  will  be  =  a -~  -f 

—  =;=  —  -I-  —  ■=   17.        But    if   the    greater   number 
2a  2  2a  ^ 

had  been  first  made  the  object  of  our  inquiry,  or  had  been 

put  =  x^  the  lesser  would  have  been  a  —  x^  and  its  square 

a^  —  2ax  +  x^^  which,  subtracted  from  x^^  leaves  2ax  —  cr 

h  n 

=  b  ;  whence  2ax  =  ^  4-  a^,  and  »r  = 1-  —  =  1 7,  the 

same  a3  before. 

PROBLEM  XXX. 

If  one  agent  A,  alone^  can  produce  an  effect  e,  in  the 
time  a,  and  another  agent  B,  alone^  in  the  time  b  ;  in  how 
long  time  will  they  both  together  produce  the  same  effect P 

Let  the  time  sought  be  denoted  by  .v,  and  it  will  be, 

ex 
as  a  :  ^  :  :  e  :  — ,  the  part  of  the  effect  produced  by  A  : 

(iheor.  3,  p.  72)  also,  vis  b  :  x  :  :  e  :  ---,  the  part  pro- 

ex      ex 

duced  by  B  :  therefore  —  -f  —  =  6'.     Divide  the  whole 

a        b 

by  e^  and  you  will  have 1 =-  1  ;     and  this,  re- 

a  b 

duced,  gives  x  =   .       After  the  same  manner,  if 

a  -^  b 

there  be  three  agents.  A,  B,  and  C,  the  time  wherein  they 

will  altogether  produce  the  given  effect  will  come  out  = 

abc 

ab  •\-  ac  '\'  be 

Example.  Suppose  A,  alone,  can  perform  a  piece  of 
work  in  10  days ;  B,  alone,  in  12  days  ;  and  C,  alone, 
in   16   days:     then  all  three  together  will  perform  the 


92  The  Application  of  Algebrb. 

same  piece  of  work  in  4/^  days  ;  for,  in  this  case,  a  being 

=  10,  ^  =  12,  c  =  16,  it  is  plain  that r- 

ab  +  ac  +  he 

, 10  X  12  X  16  -)  =  4  4 

MO  X  12  4-  10  X  16  +  12  X  16"^         ^^* 

PROBLEM  XXXL 

Txvo  travellers^  A  a?id  B,  set  out  together  from  the  same 
place^  and  travel  both  the  same  -way  ;  A  goes  28  miles  the 
first  day^  26  the  second^  24  the  third^  and  so  on^  decreasing 
two  miles  every  day ;  but  B  travels  uniformly  20  miles 
every  day  :  noxv  it  is  required  to  find  how  many  miles  each 
person  must  travel  before  B  comes  up  again  with  A  ? 

Let  X  =  the  number  of  days  in  which  B  overtakes 
A  :  then  the  miles  travelled  by  B,  in  that  time,  will  be 
20^  ;  and  those  travelled  by  A,  28  +  26  +  24  +  22, 
&f  c.  continued  to  x  terms ;  where  the  last  term  (by 
section  10,  theorem  3)  will  be  equal  to  28  —  2  X  -'^  —  1^ 
or  30  —  2x ;  and  therefore  the  sum  of  the  whole  pro- 
gression equal  to  28  +  30  —  2x  x  |^,  or  2to  — x^  (by 
theorem  4).  Hence  we  have  20a:  =  29x  —  x^  ;  whence 
20  =  29  —  x^  and  x=z9'.  therefore  20  X  9  =  180  is  the 
distance  which  was  to  be  found. 

PROBLEM  XXXII. 

To  find  three  numbers^  so  that  ^  of  the  firsts  -i  of  the  se- 
cond^ and  \  of  the  third  shall  be  equal  to  62  ;  ^  of  the 
lirst^  i  of  the  second^  and  -|  of  the  thirds  equal  to  47 ; 
and\  of  the  first  ^  \  of  the  second^  and  ^  of  the  thirds  equal 
to  38. 

Put  «  =  62,  ^  =  47,  and  c  ==  38,  and  let  the  numbers 
sought  be  denoted  by  x,  ?/,  and  z ;  then  the  conditions  of 
the  problem,  expressed  in  algebraic  terms,  will  stand 
thus  : 


X 

+ 

y. 

-f- 

z 

-_ 

a 

2 

3 

4 

X 

V 

z 

+ 

+ 

___ 

zz: 

h 

3 

4 

5 

X 

V 

z 

+ 

+ 

— — 

=: 

c, 

4 

6 

b 

to  the  Resolution  of  Problems,  ^3 

Which,  cleared  of  fractions,  become 
6;c'+    4^+    32  =  12a, 
20x  4-  \5y  +  122  =  60/^, 
I5x  +  t2?j  +  102  =  60c. 
And,    by  subtracting  the  second  of   these    equations 
from  the  quadruple  of  the  first,  in  order  to  exterminate 
2,  we  have  4x  -{-  y  =^  4^S  a —-  60b  ;   moreover,  by  taking 
3  times  the  third  from  10  times  the  first,  we  have  15x  -^ 
4z/  =  120a  —  180c;    this,  subtracted  from  4  times  the 
last,  leaves  x  =  72a  —  240^  -f  1 80c  =  24 ;     whence 

y  (48a  — 60/^  —  4v)  =  60,  and  2  (Hi-=-^^LZli^) 

o 

=  120. 

"24       60       120 

--  +  _  + =  12  +  20  +  30  =  62, 

2   ^  3   ^    4  ^        ^  ' 

24       60       120 
For<(-j+-j  +  —  =    8  +  l5+24  =  4r, 

j^  +  ^^  +  ^=    6  +  12+20=38. 
L  4         5  6 

PROBLEM  XXXIII. 

A  gentleman  left  a  sum  of  money  to  be  divided  among 
four  servants^  so  that  the  share  of  the  first  was  \  of  the  sum 
cfthe  shares  of  the  other  three  ;  the  share  of  the  second  \  of 
the  sum  of  the  other  three  /  and  the  share  of  the  third  i  of 
the  sum  of  the  other  three  ;  and  it  was  also  found  that  the 
share  of  the  first  exceeded  that  of  the  last  by  14/,  .•  the  ques- 
tion is^  rvhat  Was  the  zuhole  sum^  and  what  was  the  share  of 
each  person  ? 

Let  the  shares  be  represented  by  x^  z/,  2,  and  u^  respec- 
tively, and  let  a  =:  14;  then,  by  the  question^  we  shall 
have 

^_y±z±u 


y 


4 

u  -=.  X  -^  a* 


-          2          ' 

X  +  2  -\-U 

3 

_x  +y  +u 

y4  Ths  Amplication  of  Algebra 

Which  equations,  cleared  of  fractions,,  become 

^y  :=X  -j-Z  -^Uy 
4fZ=::  X  -^y  +Uy 

W^z  x-^a. 

Now,  if  X  be  added  to  the  first,  y  to  the  second,  and 

z  to.  the  third,  we  shall  get  (^x  -i- y  +  z  +  u)  z:z  Sx  =;:  4>y 

3x  '3jc 

=  5z ;    and  from  thence  2  =  — ,  and  ?/  =  —  ;    which 

values  being  substituted  in  the  first  equation,  we  have 

tjX  k)X  J.  KjX  .  1  t  f  1 

ax  =  —  -I- f*.  z/,  or  M  = J    but,  by  the  fourth 

4  5  20  ^     ^ 

13x 
equation,  w  =  x  r—  a;     therefore  x  ' —  a  z=,  -*-,  an^ 

X  = =  40  :  consequently  y  (— )  =  30,  z  ( — )  =  24, 

and  u  (x  —  14)  =  26  ;  and  the  whole  sum  (^x  +  y  +z 
+  u)=z  120A 


PROBLEM  XXXIV. 


Tofmdfour  iiumberSi  such  that  the  first  together  with  half 
the  second^  may  be  357  («)^  the  secoiid  zvith  ^  of  the  third 
equal  to  476  (^),  the  tiiird  with  \  of  the  fourth  equal  to 
595  (c),  and  the  fourth  w%th\  of  t^e  first  equal  to  714  (^). 

The  required  numbers  being  denoted  by  x,  y,  z,  and  w, 
and  the  conditions  of  the  question  expressed  in  algebraic 
terms,  we  have  the  four  following  equations  : 


X  -J- 

2 

+ 

a. 

y  + 

Z 
T 

= 

b. 

2  + 

u 

T 

= 

c. 

M  + 

5 

= 

d. 

to  the  Resolution  of  Probkms*  95 

From  the  first  whereof  we  get   x  =  aj  — ^  ^ ;    and 

from  the  4th,  x^Sd —  5ic  \   whence  a  —  —^sd^-^Su^ 
and  y  ^2q  —  lOd  +  lOii ;  but,  by  the  second,  y  =  b  — 

— ;  therefore  2a —  10^'+  lOw  =  ^ ,  and  z  =  Si? 

3  3 

—  6a  +  sod  —  SOii ;  but,  by  the  third,  2:  =  c ; 

whence  Sb  —  6a  +  SOd  -^  SOu  =  c  ~  — ,  and  123  — 

24a    +  120^  —   120m  =  4c  —  ic ;    consequently  u  = 

126  —  24a  -f  120^/  —4c      ^^^         ,  .  u 

— —- =  676  ;  whence  z   (c  —  — >i 

lly  ^  4  ^ 

.=  426,  y(z=b  —  ^)=:  334,  and  .v  (=  a  ~  JL)  =  190. 

Otherxvise* 

Let  the  first  of  the  required  numbers  be  denoted  by  ^ 
(as  above)  ;  then,  the  sum  of  the  first  and  ^  the  second 
being  given  equal  to  a,  it  is  manifest  that  \  the  second 
must  be  equal  to  a,  minus  the  first,  that  is  =i  a  —  .v,  and 
therefore  the  second  number  =  2a  —  2x :  moreover,  the 
sum  of  the  second,  and  \  of  the  third,  being  given  =  b  ; 
It  is  likewise  evident,  that  -J  of  the  third  must  be  equal  td 
b^  mimis  the  second,  that  is  =  3  —  %a  +  2x,  and  conse- 
quently the  third  number  itself  =  Sb  —  6a  -f-  6.v  :  in  the 
same  manner  it  will  appear  that  \  of  the  fourth  number 
=  c  —  Sb  +  6a  —  6x  ',  and  consequently  the  fourth 
number  itself  =  4c  =  123  +  24a  —  24.T  :   whence,  by  tlw 

question^  4c  —  123  +  24a  —  24,r  -j =  o^,  and  therefore 

5 

—  5t/  +  20c  —  603  4.  120a      ,  ^^  , 

^  = ^ — -^pj^ 1: =  190  ;  as  above.     ^ 

PROBLEM  XXXV. 

To  divide  the  number  90  (a)  into  four  such  parts ^  that 
if  the  first  be  increased  by  5  (3),  the  second  decreased  by 

4  (c),  the  third  multiplied  by  3  (/),  and  the  fourth  dz- 


96  The  Application  of  Algebra 

vided  by  2  (^),  the  result^  in  each  case^  shall  be  exactly  the 
same. 

Let  X,  2/,  2,  and  u  be  the  parts  required ;  then,  by  the 
question^  we  shall  have  these  equations,  viz. 
x  +  y  +  z+u=:a^  and 

X  -{-  b  =:  y  —  c  =:  dz  = — . 
e 

Whence,   by  comparing    dz  with  each  of   the  three 

other  equal  values,  successively,  x  =t  dz  —  b^  y  ■=:  dz 

-f  c,  and  It  =  dez'y     all    which   being    substituted    for 

their  equals,  in  the  first  equation,  we  thence  get  dz  — 

b-\-dz+c  +  z+  dez  =  a ;  whence  dez  -f  2dz  +  z 

z=z  a  4-  b  —  c,  and  z  =  — = -—, =   7.      There- 

^  '  de  +  2d+l 

fore  X  (^z=:  dz  —  b)  =  16;  y  (=  dz  +  c)  =  25  ;  and 

?^  (  =  dez  )  =  42. 

PROBLEM  XXXVL 

If  A  a7id  B  together^  can  perform  a  piece  of  tvork  in 
8  (a)  days;  A  aiid  C  together  in  9  (J?)  days^  and  B  and 
C  i/z  10  (c)  days;  how  many  days  will  it  take  each  person^ 
alone^  to  perform  the  same  work  ? 

Let  the  three  numbers  sought  be  represented  by  x^ 
2/,  and  2,  respectively:  then  it  will  be,  as  x  (days):  a 

(days)   :   :   1,  the  whole  work,  :  — ,    the   part    thereof 
performed  by  A  in  a  days ;  and,  as  z/  :  a  :  :  1  :  — ,    the 

y 

part  performed  by  B,  in  the  same  time ;  whence,  by  the 

question^ f- — -  ==  1    (the    whole    work).      And,    by 

X  ^      y 

proceeding   in  the  very  same  manner,    we    shall    have 

,  •  ,  .  .       b  b  ^ 

these  two  other  equations,  viz.  -—  +  —  =     1>       and 

c  c 

1 =  1 :   let  the  first  of  these  three  equations  be 

y        ^ 

divided  by  <2,  the  second  by  bj  and  the  third  by  c^  andl 

vou  will  then  have 


to  the  Resolution  of  Problems ».  9r 

L  +  L-JL 

X        y        a 

JL      i-  — -1 

X        X         b"* 

y        z        c 
which  added  all  together,  and  the  sum  divided  by  2,  give 

1 1 =  -r-H — r-l ;      fro«i    whence    each  of 

X        y        z       ^a^  2b^2c 

the  three  last  equations  being  successively  subtracted,  we 

get 

JL  —  — JLo-i.  J.  1  _  —  be  +  ac  +  ab 

_£_.J L-lJ   —  ^^~  ^c+^ 

y       2a       2b      2c  2«^c         '^ 

1  1  1  1  be    +    ac  —  ab  ^r 

—  =:  —  J-  —  —  —  —    : ; .         Hence 

X  2a         2b  2c  '^abc 

2abc  1440  ^^  , 

^  be  -{•  ac  +  ab         —  90  +  80  +  72  ^'' 

^  _  2abe  _  1440  —1723 

•^  ""  be  —  ac  +  ai^  ~  90  —  80  +  72  "■      ^"^' 
2<^^c  1440  ^^,^ 

/?>c-  +  ac  —  ab       90  +  80-/2  ^  ^' 

Otherwise* 

Let  the  work  performed  by  A  in  one  day  be  de- 
noted by  X :  then  his  work  in  a  days  will  be  ax^  and  in 
b  days  it  will  be  7»x ;  therefore  the  work  of  B  in  a  days 
will  be  1  —  ax  y  and  that  of  C,  in  b  days,  1  —  bx^  by 
the  conditions  of  the  problem  ;  whence  it  follows 
that  the  work  of  B,  in  one  day,  will  be  expressed  by 

1  '  '—  ax  '  1  — —  bx 

"— ^,  and  that  of  C,  in  one  day,  by ;    but 

the  sum  of  these  two  last  is,  by  the  question^  equal  to  — 

1         1  1 

part  of  the  whole  work,  that  is, (-  —  -—  2^^  =  —  ; 

a         b  c 


O 


98  The  Application  of  Algeh. 


ra 


,                      1.1           1        be  4-  ac  —  ab  -  ' 

whence  xz=z^+  —  —  —  =z _-- ,  equal  to  the 

2a      2b       2c  2abc  .  ^ 

work  done  by  A  in  one  day ;    by  which  divide  1   (the 

2abc 

whole),  and  the  quotient, :,  will  o-ive  the  re- 

bc  +  ac  —  ab  ° 

quired  number  of  days  in  which  he  can  finish  the  whole. 


PROBLEM  XXXVII. 

7b  fi7id  three  numbers  on  these  conditions^  that  a  times 
the  first  ^  b  times  the  second^  and  c  times  the  third^  shall  be 
equal  to  a  given  member  p  ;  that  d  thnes  the  first^  e  times 
the  second^  and f  times  the  thirds  shall  be  equal  to  another 
given  number  q ;  and  that  g  times  the  ftrst^  h  times  the 
second^  and  k  times  the  third^  shall  be  equal  to  a  third  given 
7iumber  r. 

Let  the  three  required  nun\bers  be  denoted  by  .r,  y,  and 
z-i  and  then  we  shall  have 

ax  +  bi/  +  cz  =  j&, 
dx  +  eij  -}-fz  =  q^ 
gx  +  hi/  +  y^2  =  r. 
From  d  times  the  first  of  which  subtract  a  times  the  se- 
cond, and  from  g  times  the  first  subtract  a  times  the  third, 
and  you  will  have  these  two  new  equations, 

{bdy  —  aey  +  cdz  —  afz  -^^  dp  —  aq^ 
bgy  —  ahij  +  cgz  —  akz  ^=^gp''^  ar  ; 
or,  which  are  the  same, 

bd  —  ae  X  y  +  cd —  afx  z  =:  dp  —  aqy 
and,  bg  —  ah  X  y  +  cg  —  ak  X  z  ^gp  —  ^^'• 
Multiply  the  first  of  these  two  equations  by  the  coeffi- 
cient of  y  in  the  second,  and  vice  versa^  and  let  the  last 
of  the  two  products  be  subtracted  from  the  former,  and  }'0u 
\yill  next  have  en  —  qf-  bg — ah  X  z  — bd  —  ae  X  eg — a^ 
X  Z  =:  bg  —  ah  "■■  op  —  aq  —  bd  —  ae  X  gp  —  ar;  and 
therefore  z  =  ^.y-    -'  xW^^^-bd-aexJF^^  . 

ca  —  afi  X  bg  —  ah  —  bd —  ae  X  eg  —  ak 
whence  .r  and  y  may  also  be  found* 


to  the  Resolution  of  Problems*  9^9 

^ocample.  Let  the  given  equations  be 
X  •\'    y  +      2  =  12, 
Sat  +  3z/  +    "^^  =  38, 
3;^-  +  6z/  +  102  =  83  ; 
Or,  which  is  the  same  thing,  leta=:l,^  =  l,c=:l,/7=12, 
^=  2,  e  =  3,  /=  4,  y  =  38,  ^  =  3,  A  =  6,  i  =  10,  and  r  = 
83  :    then  these  values  being  substituted   above  in  that 

.  3  —  6  X  24  —  38  —  2  —  3  X  36  —  83 

of  2,  It  will  become  === — -----    - 

2_4X    3   —  6  —2  —  3X     3  —  10 

:::::  HT =  z=.  5  X      whence,    also,    we    find 

6  —     7—1 


dp  —  aq —  cd  —  afxz^  _  24  —  38  —  2  —  4X5 
^^"~  dd—ae  ^  "^  2  —  3 


12  —  4. 


=  Zlf  =  4,  and  ;.  C=  LZi^-^^ 

—  1  '  a  .  1 

=  3. 

Having  exhibited  a  variety  of  examples  of  the  use  and 
application  of  algebra,  in  the  resolution  of  problems 
producing  simple  equations,  I  shall  now  pi'oceed  to  give 
some  instances  thereof  in  such  as  rise  to  quadratic  equa- 
tions ;  but,  first  of  all,  it  will  be  necessary  to  premise 
something,  in  general,  with  regard  to  these  kinds  of  equa- 
tions. 

It  has  been  already  observed,  that  quadratic  equations 
are  such  wherein  the  highest  power  of  the  unknown 
quantity  rises  to  two  dimensions ;  of  which  there  are 
two  sorts^  viz*  simple  quadratics,  and  adfected  ones, 
A  simple  quadratic  equation  is  that  wherein  the  square 
only  of  the  unknown  quantity  is  concerned,  as  xx  =  ab  ; 
but  an  adfected  one  is,  when  both  the  square  and  its 
toot  are  found  involved  in  different  terms  of  the  same 
equation,  as  in  the  equation  x^  +  2ax  =  bh.  The  re- 
solution of  the  first  of  these  is  performed  by  barely 
extracting  the  sqviare  root,  on  both  sides  thereof:  thus, 
in  the  equation  x^  =  ah^  the  value  of  x  is  given  =  \  ab 
(for,  if  two  quantities  be  equal,  their  square  roots  must 
necessarily  be  equal).  The  method  of  solution,  when 
the  equation  is  adfected,  is  likewise  by  extracting  the 


100  The  Application  of  Algebra 

square  root ;  but,  first  of  all,  so  much  is  to  be  added  to 
both  sides  thereof  as  to  make  that  where  the  unknown 
quantity  is  a  perfect  square  ;  this  is  usually  called  com- 
pleting the  square^  and  is  always  done  by  taking  half 
the  coefficient  of  the  single  power  of  the  unknown 
quantity  in  the  second  term,  and  squaring  it,  and 
then  adding  that  square  to  both  sides  of  the  equation. 
Thus,  in  the  equation  x$c  +  2ax  =  bb^  the  coefficient 
of  X  in  the  second  term  being  2a,  its  half  will  be  a^ 
which,  squared  and  added  to  both  sides,  gives  x"^  +  2ax 
-}-  a^  z=  b^  +  a^  ;  whereof  the  former  part  is  now  a 
perfect  square.  The  square  being  thus  completed,  its 
root  is  next  to  be  extracted ;  in  order  to  which,  it  is  to 
be  observed  that  the  root,  on  the  left-hand  side,  where 
the  unknown  quantity  stands,  is  composed  of  two 
terms  or  members ;  whereof  the  former  is  always  the 
square  root  of  the  first  term  of  the  equation,  and  the 
latter  the  half  of  the  coefficient  of  the  second  term : 
thus,  in  the  equation,  x^  +  2ax  -f-  a^  ==  b^  -f-  a^,  beforf 
us,  the  square  root  of  the  left-hand  side,  x'^  -f  2ax  +  cr.^ 
will  be  expressed  by  x  -}-  a  (for  x  +  a  X  x  -j-  a  =i  x^  + 
2ax  -f  a^).  Hence  it  is  manifest  that  x  +  a  =: 
S/b^  +  a^,  and  therefore  x  =  V/6^  -f-  c^  ' —  a;  from 
which  X  is  known.  These  kinds  of  equations,  it  is  also 
to  be  observed^  are  commonly  divided  into  three  forms, 
according  to  the  different  variations  of  the  signs :  thus 
x^  -f-  2ax  =  Z>^  is  called  an  equation  of  the  first  fori^^i ; 
^2  —  2ax  =  b^  one  of  the  second  form  ;  and  x^  —  2ax 
zzi  —  b^  one  of  the  third  form ;  but  the  method  of 
extracting  the  root,  or  finding  the  value  of  .t,  is  the 
same  in  all  three,  except  that,  in  the  last  of  them,  the 
root  of  the  known  part,  on  the  right-hand  side,  is  to 
be  expressed  with  the  double  sign  ±  before  it,  x  having 
two  different  affirmative  values  in  this  case.  The  reason 
of  which,  as  well  as  of  what  has  been  said  in  general, 
in  relation  to  these  kinds  of  equations,  will  plainly  ap- 
pear, by  considering,  that  any  square,  as  x^  —  2ax  + 
^r^  raised  from  a  binomial  root,  x  —  a  (or  a  — -  a;)  is 
composed  of  three  members ;  whereof  the  first  is  the 
square  of  the  first  term  of  the  root ;    the  second,  a  rect- 


to  the^  Resolution  of  Problems  >  101 

angle  of  the  first  into  twice  the  second ;  and  the  third, 
the  square  of  the  second :  from  whence  it  is  manifest, 
that,  if  the  first  and  second  terms  of  the  square  be  give'n 
or  expressed,  not  only  the  remaining  term,  but  the  root 
itself,  will  be  found  by  the  method  above  delivered. 

But  now,  as  to  the  ambiguity  taken  notice  of  in  the 
third  form,  where  x^  —  2ax  =  —  ^^,  or  x^  —  %ax  + 
a^  =  a^  —  ^2  .  xhQ  square  root  of  the  left-hand  side  may 
be  either  x  —  a,  or  a  —  x  (for  either  of  these,  squared, 
produces  the  same  quantity)  ;  therefore,  in  the  former  cose, 
xz=La+  Va^  —  /y^,  ?jid,  in  the  latter,  x  =  a  —  s/d^  —  h^  ; 
both  which  values  answer  the  conditions  of  the  equation. 
The  same  ambiguity  would  also  take  place  in  the  other 
forms,  were  not  the  root  (at)  confined  to  a  positive  va- 
lue. 

When  the  highest  power  of  the  unknown  quantity  hap- 
pens to  be  affected  by  a  coefficient,  the  whole  equation 
must  be  divided  by  that  coefficient ;  and  if  the  sign  of  that 
power  be  negative,  all  the  signs  must  be  changed  before 
you  set  about  to  complete  the  square. 

All  equations  whatever,  in  which  *  there  enter  only- 
two  different  dimensions  of  the  unknown  quantity,  where- 
of the  index  of  the  one  is  just  double  to  that  of  the  other, 
are  solved  like  quadratics,  by  completing  the  square : 
thus,  the  equation  x^  +  2ax'^  =  b^  by  completing  the  square, 
will  become  x^  +  2ax'^  •{-  a^  =  b  +  a^  -,  whence, 
by  extracting  the  root  on  both  sides,  x^  +  a  =  Vb  -{-  cr  ; 
therefore  y^  =z  V  b  +  a^  —  «,  and  consequently  x  = 

^Vb  +  a^  —  a. 

These  things  being  premised,  we  now  proceed  to  the  re- 
solution of  problems. 


PROBLEM.  XXXVIII. 

To  find  that  number^  to  which  20  being  addcd^  and  from 
which  10  being  subtracted^  the  square  of  the  sum^  added  to 
txvice  the  square  of  the  remainder^  shall  be  17475. 

Let  the  number  sought  be  denoted  by  x  ;  then,  by  the 


102  The  Application  of  Algebra 

conditions  of  the  question,  we  shall  have  x  +  20*]^  +  2 
X  oc  —  lOl^  =  17475  ;  that  is,  x^  +  40a:  -f  400  +  2a:^ 
.— ,  40x  +  200  =  17475;  which,  contracted,  gives  %x^ 
=  16875.  Hence  x^  =  5625 ;  and,  consequently,  x  ^ 
V5625  =  75. 

PROBLEM  XXXIX. 

To  divide  100  into  txvo  such  parts ^  that^  if  they  be  mul- 
tiplied  together y  the  product  shall  be  2100. 

Let  the  excess  of  the  greater  part  above  (50)  half  the 
number  given,  be  denoted  by  x;  then  50  +  a:  will  be 
the  greater  part,  and  50  —  x  the  lesser ;  therefore,  by 
the  question^  50  +  x  X  50  —  x,  or  2500  —  x^  =  2100  ; 
whence  x^  =  400,  and  consequently  x  =  V400  3=  20; 
therefore  50  +  x  =  70  =  the  greater  part,  and  50  —  x 
=  30  =  the  less. 

PROBLEM  XL. 

What  two  'numbers  are  those^  which  are  to  one  another  in 
the  ratio  of^  (a)  to  5  (<^),  and  whose  squares^  added  toge- 
ther, make  1666  (c).^ 

Let  the  lesser  of  the  two   required   numbers  be  x  ; 

then,  a  :   b  :  :  X  :   —  =  the  greater;     therefore,    by 
a 

the  qxiestion,  x^  -{ =  c  ;  whence  a^x^  +  b'^x^  =  aV, 


a^c  ,  / 


X 


and  x^  =  -~ — --;    consequently  .v   —  ^     „ 

a^  +  b^  ^         ^  ^  a^  +  b^ 

f        c  bx 

^  K^'iT' — r    =  21   =  lesser  number,  and  —  =  35  :=: 
^  a^  ^  6^  a 

the  greater. 

PROBLEM  XLL 

To  find  two  numbers^  whose  difference  is  8,  and  product 
240. 

If  the  lesser  number  be  denoted  by  .v,  the  greater  will 
be  A'  +  8  ;  and  so,  by  the  question,  we  shall  have  x'^  +  8.t 


tQ  the  Resolution  of  Problems*  103 

=  240.  Now,  by  completing  the  square,  x^  +  %x 
+  16  (=  240  -f  16)  =  256;  and,  by  extracting  the 
root,  X  +  4<  =  V256  =  16  :  whence  x  =  16  —  4  =  12  ; 
and  AT  +  8  =  20 ;  which  are  the  two  nijmbers  that  were 
t*o  be  found. 

PROBLEM  XLII. 

Tc  find  two  numbers  whose  difference  shall  be  12,  and 
the  sum  of  their  squares  1424. 

Let  the  lesser  be  a:,  and  then  the  greater  will  be  .r  +  12  ; 
therefore,  by  the  problem^  x  +  12*)^  +  x'^  =  1424,  or 
2x^  +  24.V  +  144  =  1424;  this,  ordered,  gives  x^  -f 
\2x  =  640 ;  which,  by  completing  the  square,  becomes 
x^  +  12;v  +  36  (=  640  +  36)  =  676  ;  whence,  extract- 
ing  the  root  on  botli  sides,  we  have  .v  +  6  =  (V676)  26  ; 
therefore  x  =  20,  and  ;^^  +  12  =  32,  are  the  two  numbers 
required. 

For  1-2-20  =12, 

^"^    132^+  202=1424. 

PROBLEM  XLIIL 

To  divide  36  into  three  such  pai'ts  that  the  second  maif 
exceed  the  first  bij  4,  a7id  that  the  sum  of  all  their  squares 
may  be  464. 

Let  X  be  the  first  part,  then  the  second  will  be  x  +  4  ; 
and,  the  sum  of  these  two  being  taken  from  (36)  the 
whole,  we  have  32  —  2x^  for  the  thircl,  or  remaining 
part ;  and  so,  by  the  question^  x'^  +  x  +  ^  +  32  —  2xY 
=  464,  that  is,  e>x^  —  120;^-  +  1040  =  464 ;  whence 
6^  _  120:v  =  —  576,  and  x^  —  20x  =  —  96.  Now, 
by  completing  the  square,  x^  —  20a:  +  100  (=  100 
—  96)  =  4  ;  and,  by  extracting  the  root,  x  - —  10  = 
^  2.  Therefore  .r  =  10  q:  2,  that  is,  ;c  =  8,  or  ;\r  = 
12  ;  so  that  8, 12,  and  16  are  the  three  numbers  require^. 

PROBLEM  XLIV. 

To  divide  the  number  100  (a)  into  two  such  parts  that 
their  product  and  the  difference  of  their  squares  7nay  be 
equal  to  each  other. 


104  The  Application  of  Algebra 

L^t  the  lesser  part  be  denoted  by  x\  then  the  greater 
will  be  a  — ^oc^  and  we  shall  have  a  —  at  x  a.  =  a  —  ~x^ 
—  x^^  that  is,  ax^^x^  :=:  (^  ' —  2«.r  ;  whence  x^  •—  2>ax 
=:  —  a^  .     and,   by  completing  the  square,    x^  —  Zcpc 

+  —    =  ( —        "^  "^-^   "dT '    ^    which  the  root  being 

extracted,  there  comes  out  x ~  ±  \/-_^,  and 

2  ^4 

therefore  ;^  =  —  ±  W_^.      But  at,  by  the  nature  of 

the  problem,  being  less  than  «,  the  upper  sign  (+)  gives 

X  too  great ;  so  that  a;  =  -^ y— ^  =  38,19658,  fcPc. 

must  be  the  true  value  required. 

PROBLEM  XLV. 

The  sum^  and  the  sum  of  the  squares^  of  two  numbers  be- 
ing given  ;  to  find  the  numbers. 

Let  half  the  sum  of  the  two  numbers  be  denoted  by  «, 
half  the  sum  of  their  squares  by  ^,  and  half  the  difference 
of  the  numbers  by  x  ;  then  will  the  numbers  themselves 
be  represented  by  a  —  at,  and  a  •\-  x^  and  their  squares 
l^y  ^2  —  2flfx  +  x^^  and  c^  +  'Hax  +  a:^  ;  and  so  we 
have  c^  —  2<2Ar  +  x^  +  c^  +  "^ax  +  x^  =z  2^,  bi/  the  ques- 
tion. Which  equation,  contracted  and  divided  by  2,  gives 
a^  ^  x^  ^  b  ;  whence  x^  z=  b  —  a-,  and  consequently  x  = 
V^  —  a^-  Therefore  the  numbers  sought  are  a  —  Vb — a^, 
and  a  +  Vb  —  a^» 

PROBLEM  XLVL 

The  sumy  and  the  sum  of  the  cubes  of  two  numbers  being 
given  ;  to  find  the  numbers* 

Let  the  two  numbers  be  expressed  as  in  the  preceding 
problem,  and  let  the  sum  of  their  cubes  be  denoted  by 
c.  I'herefore  will  a  —  .r]''  +  a  +  xf  =  c,  that  is,  by 
involution    and    reduction,  2a^  +  6ax^  =  cj     whence 


to  the  Resolution  of  Problems •  105 

iQUX^  ■=•  c  —  2«^,  vT^  =      "T  —  = ,  and  .v  ==: 

6a  6a  3' 

PROBLEM  XLVII. 

The  sum^  and  the  sum  of  the  biquads'ates  {or  4thfowers) 
of  two  numbers  being  given  ;  to  find  the  numbers. 

The  numbers  being  denoted  as  above,  we  shall  liere 
have  a  —  x\  +  a  -f  a;]^  =  ^,  that  is,  Sa^  +  I'la^x^  + 
2;c^  =  d\  from  which,  by  transposition  and  division, 
x^  +  ^(j^x^  =  i^  —  «"* ;  and,  b}'-  completing  the  square, 
x^  -f.  6a2x2  -f-  9a4  =  |^  +  Sa^  ;     whence   x'^  -f  Sa^  = 

V^^  +  8a^  ;  and,  consequently,  x  =  \ — Sa^  +  \^\d  +  8a^ 

PROBLEM  XLVIII. 

The  swn^  and  the  sum  of  the  5th  powers  of  two  numbers 
bei7ig  given  ;  to  find  the  numbers* 

The    notation  in  the    preceding    problems  being  still 

retained,  we  shall  have  2a^  +  SOa^;^^  +  \Oax'^-=.e\  and 

e  o^ 

therefore  x^  -f  ^c^x^  = ;     and  x'^  -f-  ci^  = 

10a 5j_ ^ 

v/ 1 J  whence  ^  =  \f  V f- a^. 

^\Qa       S  ^      lOa^    5 

PROBLEM  XLIX. 

What  two  numbers  are  those^  whose  product  is  120  {a)^ 
and  if  the  greater  be  increased  by  8  {b\  and  the  lesser  by 
5  (c),  the  product  of  the  txvo  numbers  thence  arising  shall 
be  300  (d)  P 

If  the  greater  number  be  denoted  by  ^,  and  the  lesser 
by  t/,  we  shall  have 
xy  =  a,  and 

X  -j-b  X  y  -i-  c  =  //,  by  the  conditions  of  the  question- 
Subtract  the  first  of  these  equations  from  the  second,  and 
you  will  have  x  +  b  x  y  -{-  ^  —  xy  =  d  —  a,  that  isy 
ex  +  by  +  be  =  d  —  a;    where  both  sides  being  multi- 

P 


106  The  Appltc<ption  of  Algebra 

plied  by  x  (in  order  to  exterminate  t/),  we  thence  have 
cx^  -f-  hxij  4-  hex  •=:  dx  -—^  ax  ;  but  xy  being  =  a,  there- 
fore is  hxy  =  ah^  and  consequently,  by  substituting  this 
value  in  the  last  equation,  cx^  j^  ah  -{•  hex  =  dx  —  ax  i 
whence  cx^  +  hex  +  ax  —  dx  =  —  ah  ;  and  therefore 

c^^  +  hx  -j =  ' ;    which,  by  makinj^  f  ^. 

c  c  c  ^ 

h  (==  28),  will  become  x^  — >  fx  = ; 

c  c  c 

hence  x^  ~fx  +  ip  =  _  ^  +  i/^,  x-if=± 

\fi/^-  ^\  and  X  =  If  ±sl^p  -  7  =  !«• 
or  =  12  ;  and  consequently  y  ( — )  =  10,  or  =  7-^5. 

For  |liiLl2--=i£2_     "^     • 

(.  12  +    8  X  10  +  5  =  300. 
(.16  +    8  X  7,5  +  5  =  300. 

PROBLEM  L. 

To  find  tnvo  niimhers^  such  that  their  sura^  thtlr  product^ 
and  the  difference  of  their  squares^  may  he  all  equal  to  one 
another. 

The  greater  being  denoted  by  x^  and  the  lesser  by  y^ 
we  have  x  +  ?/  =  xy^  and  x  -\-y  =.  x'^  —  y'^  \  the  last  of 
these  equations,  divided  by  x  -^  y^  gives  1  =  .v  - —  y  , 
whence  ^  =  1  +  ^  ;  this  value,  substituted  for  x  in  the 
first  equation,  gives  1  +  2z/  =  z/  -f  z/^  ;  therefore  if  —  y 
=:JU  and  y  z=z^  +V  ^  ;  consequently,  x  (1  +  y)  =  -|-  -f- 

PROBLEM  LL 

To  divide  the  numher  100  {a)  into  two  such  parts^  that 
the  sum  of  their  square  roots  may  he  14  (/;). 

Let  the  greater  part  be  x^  and  the  lesser  will  he  a  —  :v  ; 
therefore,  hy  the  prohkm^  V  x   +  Va  ~  x  =  h;    and. 


tt  the  Resolution  ofFroblems*  107 


by  squaring  both  sides,  x  +  2Vax  >-^  xx  +  a  —  x  ■=.  bb  -, 
whence,    by  transposition  and  division,  \^ax  —  xx  ^=' 

— — — :     therefore,    by  squaring   again,    ax  —  xx  = 


L   or  x^  —  ax  z=:  ( L-)  = L 

b^a  a"  ,  a      ^       \  2ah'  ~  b''  a 

^ —  — ,    and  .r  =  -— -  +  v =  —  4- 

2  4'  2^>4  2^ 

-—  V2a  —  <^2  __  54  _.  |.|^^  greater  part ;  whence  a  —  .r  ==: 
36  =  the  lesser  part* 

PROBLEM  LII. 

A  grazier  purchased  as  mayiij  sheep  as  cost  hrni  60/.  bul: 
ofzvliich  he  reserved  15^  and  sold  the  remainder  for  54/. 
and  gained  tiw)  shillings  a-head  by  them :  the  question  is, 
hoxv  jnamj  sheep  did  he  buy^  and  xvhat  did  they  cost  him  a- 
liead? 

Let  the  number  of  sheep  be  x\  then  if  1200,  the 
number  of  shillings  which  they  all  cost,  be  divided  by  at, 

the  quotient,   ,  will,    it  is  evident,  be  the  number 

X 

of  shillings  which  they  cost  him   a-piece  ;     and  so  the 
number  of   shillings    they  were    sold  at  per   head  will 

be [-  2,  by  the  question  ;    and  therefore  this,  mul- 

tiplied  by  x  —  15,  the  number  of  sheep  so  sold,  will  giye 

1200  +  2.V 30,  equal  to  the  whole  number 

X 

@f  shillings  which  they  were  all  sold  for;    that  is,  11/0 
+  2:^  — .1522-  =  1080:    hence  we  have  liro.r  +  2a^ 

X 

—  18000  =  1080;i\  2x'^  +  90.V  =  18000,  x^  +  45.r  =r 
9000,  and  x  =  V9506.25  —  22.5  ■=.  75,  the  number  of 

1200 

sheep  ;  and  conseqnentlv rr^  1 6  shillings,  the  price  ©f 

/  5 
each. 


108  The  Application  of  Algebra 


PROBLEM  LIIL 

Trvo  country 'Xvomen^  A  and  B,  betwixt  them^  brought  100 
(c)  eggs  to  market;  they  both  received  the  same  sum  for  their 
eggs  ;  but  A,  who  had  the  largest  and  best^  says  to  B,  Had 
I  brought  as  many  eggs  as  you^  I  should  have  received 
18  (a) pence  for  them;  butj  replies  B,  had  I  brought  no 
more  than  you^  I  should  have  received  only  8  (hi)  pence  for 
mine:  the  question  is^  to fnd how  many  eggs  each  person 
had. 

If  the  number  of  eggs  which  A  had  be  =  x^  the 
number  of  B's  eggs  will  be  =  c  —  x ;  therefore,  by  the 

problem^  it  will  be,  c  —  x  i  a  i  x  x  \  ^  =  the  num- 
ber of  pence  which  A  received ;  and  as  .r  :  ^  :  :  c  —  x  i 
b  X  c  —  X 


X 


the  number  of  pence  which  B  received : 


ax          b  ^  c  I      X 
whence,  again,  by  the  problem^  — '■ —  = ;   and 

c  •— —  X  X 

therefore  ax'^  =  b  x  c  —  xY  =  bc^  —  2bcx  +  bx^ ; 

^bcx              bc^ 
which  equation,  ordered,  gives  x^  •\ — ^- — ■,  =  •  ; 

a  —  o        a  ——  b 

from  whence  x  comes  out  (  =  v >    + 

^        ^  a  —  b 


.^ )  ^ . —  =  40.      But  the  value  of  x  may 

a  —  b  a  —  b 

be  otherwise  more  readily  derived  from  the  equation 
ax^  =  b  X  c  —  xY',  without  the  trouble  of  completing 
the  square ;  for  the  square  root  being  extracted  on  both 
sides  thereof,  we  have  xV  a   =c  —  x  X  \^  b  ;  whence 


x\/a  +  x\/b=zcx/b^  and  consequently  x 


_      c  Vi 


\/  a  +  '\/  h 

100^/8  100  V  4  .^         ,   r 

: = =  40,  as  before* 

V  18  +  V8      \/9  +\/4  "^ 


to  the  Resolution  of  Problems.  109 

PROBLEM  LIV. 

One  bought  \20  pounds  of  pepper^  and  as  inaJiy  ofgingei^ 
and  had  one  pound  of  ginger  more  for  a  crown  than  of  pep- 
per; and  the  whole  price  of  the  pepper  exceeded  that  of  the 
ginger  by  six  crowns :  how  many  pounds  of  pepper  had  he 
for  a  croxvuj  and  how  many  of  ginger  ? 

Let  the  number  of  pounds  of  pepper  which  he  had 
for  a  crown  be  x^  and  the  number  of  pounds  of  ginger 
will  be  ;^  +  1  ;  moreover,  the  whole  price  of  the  pep- 
.„   ,       120  ,1  r    i_         •  120 

per  will  be  —   crowns,  and  that  ot  the  gmger ; 

'^  X  X  +  X 

therefore,  by  the  question^  —  —  =  6 ;  whence 

'      ^  ^  ^      X  X  '\-l 

120a:  +  120  —  120.r  =:  6:^^  +  6^,  and  therefore  x^  +  x 
=  20  j  which,  solved,  gives  ;r  =  4  =  the  pounds  of  pep- 
per, and  X  +  1  z=z  5  =  those  of  ginger, 

PROBLEM  LV. 

To  find  three  numbers  in  arithmetical  progressio7i^xvhere- 
of  the  sum  of  the  squares  shall  be  1232  (a),  and  the  square 
of  the  mean  greater  than  the  product  of  the  two  extremes  by 
16  (^). 

Let  the  mean  be  denoted  by  x^  and  the  common  dif- 
ference by  y  ;  then  the  numbers  themselves  will  hQ  x  —  i/, 
AT,  and  X  +  y  \  and  so,  by  the  problem^  we  shall  have  these 
two  equations : 

X  —  yl  -f  y^  +  X  4-  2/^  =  a^  and 

x^  •=^  X  —  y  X  X  +  y  +  b:  these,  contracted,  become 
Sx^  +  2y^  =  a,  and  x^  =  x^  —  y^  +  b ;  from  the  latter 
whereof  we  get  y^  =  b  =  16;  and  consequently  y  = 
V  ^  =  4 ;  which,  substituted  for  y  in  the  former,  gives 

Zx^  •}-  2b  =z  a;    whence  x^  =  ^ ,  and  therefore 


F 

.r  =  y  — 


—  =  20  ;   so  that  the  three  required  numbers 


3 
are  16,  20,  and  24. 

For  /  ^^^  +  ^^^  +  24^  =  1232. 
"'"^'^    t202--16   X24  =16. 


1 1 .0  7 '//  e  Application  of  Algebra 

PROBLEM  LVL 

To  fmd  two  numbers  whose  difference  shall  he  10  (a), 
and  if  ^00  (Ji)  he  divided  by  each  ofthem^  the  difference  of 
the  quotients  shall  also  be  equal  to  10  («). 

The  lesser  number  being  represented  by  :c^  the  greater 
will  be  represented  by  ^  +  a;  and  therefore^  by  th-:  pro- 
blem^   =  a ;     which,  freed  from  fractions, 

X         oc  -^  a 

gives  hx  +  ha  -^  bx  •=.  ax^  +  a^x^  that  is,  ba  =  ax^  + 
a^x  y  whence,  dividing  by  a,  and  completing  the  square^ 
we  have  x^  +  ax  •}-  \a^  :=.  b  +  \a^  ;  therefore  x  +  ^a:=: 
\/h  ^  i-a^,  and  consequently  x  =  \^b  +  \(j^  —  i^  =  20, 
the  lesser  number  :  whence  x  •\-a^=.  30,  the  greater  num- 
ben 

PROBLEM  LVIL 

To  find  two  numbers  whose  sum  is  80  («),  and  if  they 
bt  divided  alternately  by  each  other  ^  the  sum  of  the  quotients 
shall  be  ^  (b). 

If  one  of  the  numbers  be  .r,  the  other  will  be  «  —  Xj 

X            a  t      X 
and  we  shall  therefore  have -f-  =  b  :  which 

a  —  X  X 

equation,  brought  out  of  fractions,  becomes  x'^  +  a^  — 
^ax  +  x"^  =  abx  —  bx^ ;  and  this,  by  transposition, 
gives  2x'^  +  bx^  —  2ax  —  abx  =  —  a^,  that  is, 
2  +  6  X  c^'^  —  2  -f.  6  X  cix  =  —  a^  ;  whereof  both  sides 
being    divided    by  2   +  6,    we    have    x^  —  ax    =  ■ — 

a^ 
;    whence,  by  completing  the  square,  x^  —  ax  -f. 

a^         a^  ft2  p  ^— 

—  = , ;   hence  x  —  la  =  ±  V j, 

4  4  2  -\-  b  ^  4<        a  +  b^ 

if  m  fl  ft 

and  X  ^  —  ±  V—  — =  60,  or  =  20  ;  which  two 

2       >  4        2  +  /^  ' 

are  the  numbers  that  were  to  be  found. 
PROBLEM  LVIIL 

,  To  divide  the  number  134  {a)  into  three  such  parts ^ 
that  once  the  ^first^  txvice  the  second^  and  three  times  the 


to  the  Resolution  of  Problems.  Ill 

r/iird^  added  together^  may  be  =  278  (^),  arid  that  the  su?7z 
of  the  squares  of  all  the  three  parts  may  be  =  6036  (c). 

Let  the  three  parts  be  denoted  by  x^  i/,  and  2,  respec- 
tively ;  then,  from  the  conditions  of  the  problem,  we  shall 
have  these  three  equations  : 

X  +    y  +    z  =  a^ 

X  +2y   +  3z  =  b^ 

x^  -{-  y^  +  z^  =z  c. 
Let  the  first  of  these  equations  be  subtracted  from  the 
second,  whence  y  +2z  =  b  -^  a^  or  y  =  b  —  cr  —  Sz; 
also,  if  the  double  of  the  first  be  subtracted  from  the 
second,  there  will  come  out  z  —  x  z=z  b  —  2dz,  or  x  ^z 
-f  2a  —  b:  wherefore,  iifho^  put  =z  b  —  a  (=  144), 
g  =  A  —  2a  (=  10),  and  for  y  and  .r,  their  equals/ —  2z 
and  z  —  gj  be  substituted,  our  third  equation,  x^  +  y^ 
-[-  2;^  =  c,  will  become  zz  —  2^2  +  gg  +  ff  —  4/2 
4-    422    +   22    =    c  ;      which,    ordered,    gives    2^    — 

^^  "^'^.  X  2  =  ^' ^-  ;    whence,  by  putting  h  = 

cyf    t     p.  298 

^  "^  ^   (  =  . — ),  and  completing  the  s.quare,  &f c.  2  is 
h  Ic  —  p  —  ^        ^2  149         1 

>0  :  therefore  y  {=■  f —  22)  =  44,  and  .v  (  =  2  — g)  ^ 

'0. 

PROBLEM  LIX. 

A  traveller  sets  out  from  one  city  B,  to  go  to  another  C,  at 
the  same  time  that  another  traveller  sets  out  from  Qfor  B  ; 
they  both  travel  uniformly^  and  in  such  proportion  that  the 
former^  four  hours  after  their  meetings  arrives  at  C,  and  the 
latter  at  B,  in  nine  hours  cfter  :  now^  the  question  is  to  find 
'n  how  many  hours  each  person  performed  the  journey. 

D 

B 1 . C 

Let  D  be  the  place  of  meeting,  and  put  a  =  4,  /i  =  9, 
and  X  =  the  number  of  hours  they  travel  before  they 
meet :  then,  the  distances  gone  over,  with  the  same  uni- 
form motion,  being  always  to  each  other  as  the  times  in 


112  The  Application  of  Algebra 

which  they  are  described,  we  therefore  have,  BD  : 
DC  I  I  X  (the  time  in  which  the  first  traveller  goes  the 
distance  BD)  :  «  (the  time  in  which  he  goes  the  distance 
DC)  ;  and,  for  the  same  reason,  BD  :  DC  :  \  b  (the  time 
in  which  the  second  goes  the  distance  BD)  :  x  (the  time 
in  which  he  goes  the  distance  DC)  :  wherefore,  since  it 
appears  that  x  is  to  a  in  the  ratio  of  BD  to  DC,  and  b  to 
X  in  the  same  ratio,  it  follows  that  x  i  ax  :  b  \  x;  whence 
oc^  =  ab^  and  x  =z  Vab  (=  6)  ;  therefore  a  +  Vab  =10, 
and  b  +  \'^ab  =15,  are  the  two  numbers  required. 

PROBLEM  LX. 

There  are  four  numbers  in  arithmetical  progression^ 
xvhereofthe  product  of  the  extremes  is  3250  («),  and  that 
.  of  the  means  3300  (Ji)  :  what  are  the  numbers? 

Let  the  lesser  extreme  be  represented  by  z/,  and  the 
common  difference  by  x ;  then  the  four  required  numbers 
will  be  expressed  hy  y^y  +  x,  y  +  2x^  and  z/  +  3x  :  there- 
fore, by  the  question j  we  have  these  two  equations,  viz. 
y  X  y  +  3a%  or  2/2  +  3xy  =  a,  and 

y  ^  X  X  y  +  2x^  or  2/2  +  Sxy  -f  2x^  =  b ;  whereof 
the  former  being  taken  from  the  latter,  we  get  2x^  = 

b  —  a:  and  from  thence  x  =  j^ _II1~.   =  5       But,   to 

find  y  from  hence,  we  have  given  y^  +  3xy  =  a  (by  the 
first  step)  ;  therefore,  by  completing  the  square,  £5fc.  y  = 

a  4-  —  . =  50 :  and  so  the  four  numbers  are  50, 

4  2 

35^  60,  and  65. 

PROBLEM  LXL 


4 


The  sum  (30)  and  the  sum  of  the  squares  (308)  of  three 
numbers  in  arithmetical  progression  being  given  ;  to  find 
the  numbers. 

Let  the  sum  of  the  numbers  be  represented  by  3^, 
.the  sum  of  their  squares  by  c,  and  the  common  diffe- 
rence by  X :    tlien,  since  the  middle  term,  or  number, 


to  the  Resolution  of  Problems*  1 1<3 

from  the  nature  of  the  progression,  is  =  b^  or  |  of  the 
whole  sum,  the  least  term,  it  is  evident,  will  be  ex- 
pressed hj  b  —  :r,  and  the  greatest  by  b  -^  x  \  and 
therefore,  by  the  queMwn^  we  have  this  equation, 
b  —  v"]2  +  b^  +  b  +  x\^  =  c;  which,  contracted, 
gives  3b^  +  2:>c^    =  c ;     whence  2x^   =  c  —   ob^^    and 

x=:  JiZZ^  =  ^'  Therefore  8,  10,  and  12,  are  the 
three  numbers  sought. 


PROBLEM  LXII. 

Hmnng  giveii  the  sitm  (^),  and  the  sum  of  the  square^ 
(c),  of  any  giueri  number  af  terms  in  arithmetical  progreS' 
sion;  to  find  the  progression. 

Let  the  common  difference  be  ^,  the  first  term  x  -f  e, 
and  the  number  of  terms  n :  then,  by  the  question^  we 
shall  have 

X  '{'  €  +  X  -\-  Ze  -\-  X  '\'  :^e X  A-  ne  ^  b^  and 

X  4-  e'Y^  X  -f  '^e^-^  x  -|-  Se]^ x  -f-  ne]^  =  c. 

But  (by  section  10,  theo.  4)  the  sum  of  the  first  of  these 

progressions  is  nx  -j — 1— '- —  :     and  the  sum  of  the 

second     (as  will  be  shown  further  on)     is    =    7ix^    + 

7i»n+  1  .  xe  + -^- :: ■ :  therefore  oirt^ 

two  equations  will  become 

nx  -f     . i- =  b,  and 

2 


71  -4-  1  .     2n  -f-   1  .  e?^ 


nx^  +  21  •  n  -{-  i  .  xe  + :•- =  o 

o 

Let  the  former  whereof  be  squared,  and  the  latter  muK 

ti plied  by  7Z,  and  we  shall  thence  have 

nV  +  7z2  .'^T+T.xe  +  ^.^'•^^  +  0''"'  ^  ^2^  and 

4 


'2  W2  -f-  1  .  2'i   -h  1  •  C 


n^X^    H>   72^  .  n    +    1  .  ;c^    +     ^\'^^^^'r!.J^ =:    ^^C 


114  7  lie  Applkatioji  of  Algebra 

let  the  first  of  these  be  subtracted  from  the  second,   so 

shall — T • =  nc—b^* 

But  ^^^ '  ^  +  ^ '  ^^  +  1  __  y^^ » y^  + 1] Ms  =?i^  ,¥Ti  X 

6  4 


2;z  -f-  1         72  +  1  2      r-T   V     8^i  +  4^  —  67Z  —  6 

6  4  ^  24 


72    +  1  . 


2/2 2  __  ^^^  •  ^  +  1   •  ^2  —  1 ^^^  •  72^  — ^  1  : 

24      "^"         "^12  12 

Therefore  ^i-l-^^^ •-^~  =  ?2c   —    b\    and     ^    = 

lt:^nc—12¥  (b        72+1   .  g\ 

\ -J         .  ;  whence  .r  (  -;  — — /    is  known- 

Example :  Let  the  given  number  of  terms  be  6,  their 
sum  33^  and  the  sum  of  their  squares  199  ;  then,  by 
writing  these  numbers,  respectively,  for  72,  ^,  and  c,  we 
shall  have  e  =  1  ;  whence  .v  =  2,  and  the  required 
numbers  3,  4,  5,  6,  7,  and  8. 


PROBLEM  LXIIL 

Two  post-boys^  A  ^tz^  B,  set  oiit^  at  the  same  time^from 
tivo  cities  500  7niles  asunder^  in  order  to  7neet  each  other  :  A 
rides  60  7niles  the  first  daij^  55  the  second^  50  the  third^  and 
so  on^  decreasing  5  miles  every  day  :  bjit  B  goes  40  miles  the 
first  day^  45  the  second^  50  the  third^  &c.  increasing  5 
miles  every  day  ;  7ioxv  it  is  required  to  find  in  xvhat  number 
of  days  they  xvill  meet. 

In  order  to  have  a  general  solution  of  this  problem, 
let  the  first  day's  distance  of  the  post  A  be  put  =  m^ 
and  the  distance  which  he  falls  short  each  day  of  the 
preceding  =  ^y  also  the  first  day's  distance  of  the  pqst 
B  =  /;,  and  the  distance  which  he  gains  each  day  :=.  e ; 
and  let  x  be  the  required  number  of  days  in  which  they 
meet :  then  the  whole  distance  travelled  by  A  will  be 
expressed  by  the  following  arithmetical  progression  : 
m  +  m  —  d  +  m  —  2d  +  m  —  3d^  &c.  and  that  of  B  by 


to  the  Resolution  of  Problems.  1 1 5 


p+p  +  e+p  +  2e+p  +  3^,  &c.  where  each  pro- 
gression is  to  be  continvied  to  x  terms.  But  the  sum  of 
the  first  of  these  progressions  {by  Sect.  10,  ^/zeor.  4)  is  = 

'^ix  —  ^iLlZliil-.,  and  that  of  the  second  =  px  + 


^'  ^  ^ -— :  therefore  these  two  last  expressions,  add- 
ed together,  must,  by  the  conditions  of  the  question, 
be  equal  to  500  miles,  the  whole  given  distance  ;  which 
we  will  call  ^,  and  then  we  shall  have  p  -^  rn  X  x  + 

XXX  —  IX  e  —  d        ,          ^     ^   g-x  X  x—1         ,    , 
5 =  ^  orfx  +  ^ =  b,  b) 

writing  y=  p  +  w,  and  ^  =  <?  —  d;  which  equation  is 
reduced    to   ^x^  —  g-x  +  2fx  =  2b^    or    x^  —  x  + 
2fx         2b 
■^    z=z  -^;     whence,  by  completing  the  square,  ^c.  x 

^/ii        }         iV         f        1        ^       . 
comes  out  =  V 1- —  —  -I •      But,  m 

the  particular  case  proposed,  the  answer  is  more  snnple, 
and  may  be  more  easily  derived  from  the  first  equation 


XXX  —  1  X  e  —  d 


p  +  mx  X  -i —  ^  ;  for,  e  being  =  d, 


'_ ^yill  \iQYe  entirely  vanish  out  of  the 

equation;    and  therefore  x  will  be  barely  =  = 

■7-— — --  =  5.  The  same  conclusion  is  also  readily  de- 
40  +  60  ^  ^ 

rived,  without  algebra,  by  the^help  of  common  arith- 
metic only :  for,  seeing  the  sum  of  the  two  distances  tra- 
velled in  the  first  day  is  100  miles,  and  that  the  post  B 
increases  his  distance,  every  day,  by  just  as  much  as  the 
post  A  decreases  his,  it  is  evident,  that,  between  them 
both,  they  must  travel  100  miles  every  day  ;  therefore^ 
if  500  be  divided  by  100,  the  quotient  5  will  be  the 


116  ,  The  Application  of  Algebra 

number  of   days  in  which  they  travel   the   whole  500 
miles. 


PROBLEM  LXIV. 

Txv9  persons^  A  and  B,  set  out  together  from  the  samt 
place^  and  travel  both  the  same  rvay :  A  goes  8  ^niles  the 
first  datf^  12  the  second^  16  the  thirds  and  so  on^  increasing 
4  iniles  every  day ;  but  "S^  goes  1  mile  the  first  day^  4  the 
second^  9  the  thirds  and  so  on^  according  to  the  square  of  the 
number  of  days:  the  question  is^  to  find  how  many  days  each 
must  travel  before  B  comes  up  again  rvith  A. 

Let  (4)  the  common  difference  of  the  progression  8, 
12,  16,  £sPc.  be  put  =  e^  and  the  first  term  thereof  minus 
the  said  common  difference  =  w,  and  let  the  number  of 
terms,  or  the  days  each  person  travels,  be  expressed  by 
X :    then  the  sum  of  that  progression,  or  the  number  of 

X  X  ^  -^  1  X  c 
2 

{by  sect,  10,  theor,  4).  And  (^by  what  follows  hereafter) 
the  sum  of  the  progression  1  -f-4  +  9  .  .  >  .  y^,  or  the  dis- 
tance travelled  by  B,  will  appear  to  be  ^ — — liJt— 


miles  which  A  travels,  will  be  x  X  ni  + 


therefore,  by  the  questio7ij  we  have  


6 


\    ?:   X  -^'f"  \  X  e 
z=,  rnx  +  — ;  which,  divided  by  at,  and  con- 

.      .A       •         2;c2  4-  3^  -^  1               ^    ex  -h  e         , 
tracted,  gives =  m  -f ;    whence 

,     ,    3x  Sex         „        .    3^  1  J     I 

:ir  +  —  —  =  Sm  +  — ;     and,   by   com- 

2  2  2  2  '     -^ 

pletmg  the  square,  ^^+-~- ^T+T^  —  "7^   + 

2  2  lo  Id 

9e^    ,  Se  1  9  18^  9^2 

-     (=   3.^    +    ^   -  —    +   ^    --^    +     l6    = 

48m  4-  1   4-  6c'  -f  9e^\   _  48m  +   1   -^  Sef         , 

16  /  ""  16  '    w  enc^ 


tD  the  Resolution  of  Problems.  1 1  f 


/i   +   -.    —   -,    =   -I— L,   and   X    = 

4  *  "* 


required. 


PROBLEM  LXV. 

The  sum  of  the  squares  (a),  and  the  continual  product  (b) 
of  four  numbers  in  arithmetical  progression  being  given  ; 
to  find  the  numbers. 

Let  the  common  difference  be  denoted  by  2^,  and 
the  lesser  extreme  by  2/  —  3x  ;  then,  it  is  plain,  the  other 
three  terms  of  the  progression  will  be  expressed  by  2/  —  x^ 
y  +  x^  and  y  +  3x^  respectively  ;   and  so,  by  the  question^ 

we  have  

y  —  Z'Y  +  y  —  x'Y  4-  fj  -f  X  p  4>  y  -f-  3>x'Y  =  «^,  ani 
y  —  ctx  X  y  —  xxy  +  xxy  +  ^x=:b^ 
that  is,  by  reduction, 

4z/2  +  20:^2  =  a,  and 
y^  —  XOy^:^^  +  9x'^  =  b  ; 
from  the  former  of  which  y^  =.  \a  —  Sx"^ :    and  there- 
fore y^  =  \a^  —  I  ax^  +  25;t^^  :   these  values  being  sub- 
stituted in  the  latter,    we    have    -^^a^  —  ^ax^  +  25^;^ 

5ax^ 
—  lax^  +  50;^^*  +  9x^  =  b^  and  therefore  x"^ 

r= ;  whence,  by  completint?  the  square, 

84         16  X  84  ^     ^  16  ^5 

^4  _Sa^         25£ r  =   A  4.  _J!L_>)  =. 

84  4   X    84   X    84   ^        84         84  X    84'' 

-~ ;  therefore  x^ =  —^ ■ — ,  and  x  =r 

84    X  84  2  X  84  84  ' 


4. 


5a'   2\/8U  4.  a^ 


168 
known. 


;  whence  y  (=  V^a  —  S.r^)  is  also 


118  The  Application  of  Algebra 

PROBLEM  LXVI. 

The  difference  of  the  means  {a)^.and  the  difference  of^he 
extremes  {F)  of  four  numbers  in  continued  geometrical  pro- 
portion being  given  ;  to  find  the  numbers. 

Let  the  sum  of  the  means  be  denoted  by  x ;  then  the 

gi'eater  of  them  will  be  denoted  by  ^  "^  ^,  and  the  lesser 
by  — — -  :    whence,  by  the  nature  of  proportionals,  it 

At 

.„  -      X  •\'  a      X  —  a        X  —  a       x  —  clV 

will  be   — -—  :  — -- —  :  :  — — ~    : -L.,  the  lesser 

2  2  2  2x  +  2a ' 

X  —  a      .r  4-  a         x  ■\'  a     x  -^^  a^    , 

extreme,  and :   — • —  :  :  — ! —  :  — — — L ,  the 

'22  2        2;v— 2a' 

greater    extreme :     therefore,   by  the  problem^  we  have 

X  -^  (lY         ^  —  «T        7  1  ^1  T — 13 

! L  =r  h  ;    and  consequently  at  -f  a  I    — 

2x — la         2x  -f  2a  ^         ^  ' 

X  —  aY  =^2b  X  X  --^a  X  x  +  a^  that  is,  6x^a  +  2a^  = 

— — —  1)0^  -1-  a^ 

2b  X  ^^  —  a^ ;  whence  x^  =  -7 — -~p->  and  consequently 


'3a 


.^'  =  aJ 


^+    a 


■  Sa 

PROBLEM  LXVIL 

The  sum^  and  the  sum  of  the  squares  of  three  numbers  in 
geomet7'ical  proportion  being  given  ;  to  find  the  numbers. 

Let  the  sum  of  the  three  numbers  be  denoted  by  a, 
and  the  sum  of  their  squares  by  ^,  and  let  the  numbers 
themselves  be  denoted  by  x^y^  and  2  :  then  we  shall  have 
X  +y  +z  =:a, 
x'^  +  y^  +z^=zby 
and  xz  =  z/^. 
Transpose  .2/    in    the    first  equation,    and    square    both 
sides,  so  shall  x"^  +  2xz  +  2^  =  a^  —  2ay  +  y^  ;    from 
whei>ce»  subtracting  the  second  equation,  we  have  2^2 
—  2^2  ^  ^^2  —  2ay  +  y^  —  b:    but,  by  the  third,  2xz 
^  2?f  ;  therefore  y-  z=^  e^  —  2ay  +  z/^  —  b\  and  conse- 


to  the  Resohilion  of  Problems*  119 

quently  y  =  -^-  =  ~  -  ^.       ^ow,  to  find  -x 

and  2,  t/  may  be  looked  upon  as  known  ;  and  so,  by  the 
second  eqviation,  we  have  given  x"  +  z^  =  b  —  z/^  ; 
from  whkh  subtracting  2xz  =  2z/*^,  there  arises  x^  —  2xz 
+  z^  =:b  —  3y^ ;  where,  the  square  root  being  extracted, 
we  have  cc  —  2  =  Vb  —  5y^  :  but,  by  the  first  equation, 
we  have  x  +  z  =  a  —  y ;  whence,  by  adding  and  sub- 
tracting these  last  equations,  there  results  2x  :=z  a  —  y  + 
Vb  •—  oz/z/,  and  22  =  a  —  t/  —  Vb  —  Si/y. 

PROBLEM  LXVIIL 

The  sum  Qs)^  and  the  product  (p)^  of  any  two  nujyibers 
being  given;  to  find  the  sum  of  tlie  squares^  cubes^  biqua- 
drates^  &c.  of  those  nu?nbers. 

If  thfe  two  numbers  be  denoted  by  x  and  y  ;  then  Avill 

.V  +  z/  =  5  j  ,    the  problem, 
and  xifc:ip  j    ^        ^ 
The  former  of  which,  squared,  gives  xx  +  2xy  +  yy^=i  s^ ; 
from  whence  subtracting  the  double  of  the  latter,  we  have 
x^  +  2/2  =  ^^  —  2/?,  the  sum  of  the  sqitares. 

Let  this  equation  be  multiplied  by  x  +  z/  =  5  ;  so  shall 
x^  +xy  X  X  +  y  +  y^  =  s^  —  2sp^  that  is,  x^  +p  x  s  +  y^ 
=:  s^  —  2sp  (because  xvj  = /?,  and  x  +  y  =z  s)  ;  and  there- 
fore x^  +  2/2  _-  ^3  —  3^^^  f/i^  ^^;;^  of  the  cubes. 

Multiply,  again,  by  .v  +  ^  =  ^ ;  then  will  x"^  +  xy 
X  x^  +  y'^  +  2/4  3=  ^4  —  3^2^^  ^^  x^  +  p  X  A>2  — 2p  + 
yA  __  ^^4  —  3^2^  (because  x"^  +  2/^  =  ^2  —  2p).  Conse- 
quently AT'*  +  2/*  ==  5^  —  45'2j&  +  2/j^,  tlifC  sum  of  the  biqua- 
drates. 

Hence  the  law  of  continuation  is  manifest,  being 
such,  that  the  sum  of  the  next  superior  powers  w^ill  be 
always  obtained  by  multiplying  the  sum  of  the  powers 
last  found  by  5,  and  subtracting,  from  the  product,  the 
sum  of  the  preceding  ones  multiplied  by/?.  And  thus  the 
sum  of  the  ;ith  powers,  expressed  in  a  general  i^anner^ 

wfll  be  S""  '^  72.S  ^-2^  +  n  •  ^i^^  .  s'^"'^  p^  —  n  .  ^-^^^. 


The  Appficatioii  of  Algebra 


PROBLEM  LXIX. 

The  sum  of  the  squares  (a),  and  the  excess  (^)  of  the 
product  above  the  sum  of  two  numbers  being  given  ;  to  find 
the  numbers* 

Let  the  sum  of  the  numbers  be  denoted  by  ^,  and 
their  product  by  r  ;  then  the  sum  of  their  squares  will  be 
5.2 . —  2r  {by  the  last  problem)^  and  we  shall  have  r  —  « 
=  b^  and  5^  — «  2r  =  a,  whence,  by  adding  the  double 
of  the  former  equation  to  the  latter,  s^  —  2^  =  a+2b-y 
and  consequently  s-=:\/a-\-^b'^\  +!•  From  which 
r  (z=zb  +  s)  is  likewise  known ;  and  from  thence  the 
numbers  themselves. 


PROBLEM  LXX. 

The  sum  {a) ^  and  the  sum  of  the  squares  (b)  of  four 
numbers^  in  geometrical  progression^  being  given  ;  to  find 
the  nu7nbers. 

If  X  and  y  be  taken  to  denote  the  two  middle  numbers, 
the  two  extreme  ones,  by  the  nature  of  progressionals, 

x^  Xp" 

will  be  truly  represented  by and  -^, 

y  X 

Put  the  sum  of  the  two  means  =  s^  and  their  rect- 
angle =  r;     so    shall    the    sum   of  the    two    extremes 


(XX  yy\ 
+  ^  i  be  =  a  —  5,  and  their  rectangle  also  =  r 

(by  the  nature  of  the  question).  But  (^by  problem  68) 
the  sum  of  the  squares  of  any  two  numbers  whose 
sum  is  s^  and  rectangle  r,  will  be  =  ss  —  2r  ;  and,  for 
the  same  reason,  the  sum  of  the  squares  of  our  other 
two  numbers  (whose  sum  is  a  —  ^,  and  rectangle  r)  will 
be  =  rt  —  5  p  —  2r.  Therefore,  by  adding  these  aggre- 
gates of  the  squares  of  the  means  and  extremes  together^ 
We  get  this  equation,  viz.  s^  +  «  — ^  «"]  ^  —  4r  t=r  <^. 


to  the  Resolution  of  Problems.  121 

Moreover,  from  the  equation,    —  +  liL  ^  a  —  s. 

y ^ 

we  get  x^  -{-y^  =  xy  x  o,  —  s  -=1  r  x  (i  —  s\  but  {by  the 
same  pr oh.  just  now  quoted)  x'^  +if  z=zs'^  —  Zsr  ;  therefore 

^3  —  ^8r  ■=.  ar  —  *r,  or  r  =  ;    which  vaUie  be- 

ing    substituted    for    r,    in   the  preceding  equation,  we 

have  5^  -f  a  —  ^1 =  h.     This,  solved,  gives 

'  2^  +  a  '  7  t> 

^  =  v/  — ^^^^^^^^ — I —  — '-  whence  every  thino;  else  is 

>       2       ^4aa        2a  ^  b 

readily  found. 


PROBLEM  LXXI. 

The  sum  (a)  and  the  sum  of  the  squares, (J))  of  five  iiwur 
bers^  in  geometrical  progression^  being  given  ;  to  find  thr 
numbers. 

Let  the  three  middle  numbers  be  denoted  by  x^  ?/, 

XX  22 

and  2  :  then  the  two  extreme  ones  will  be  —  and  —  ;  and 
therefore  we  shall  have 

XX  22  ^ 

4  ^  r*  %  ^^^^  question. 

^+^  +  2/^+2^  +  -,=   ^,/ 

•'  ^  .  XX  Z^ 

Put  X  +  z  =  u;    then,  by  the  first  equation, ( ^ 

y       y 

r=  a  —  u  —  y.  Wherefore,  seeing  the  sum  of  the  two 
extremes  is  expressed  by  a  —  u  —  y,  and  their  rectangle 
by  y^  (see  theor.  7,  sect.  10),  the  sum  of  their  squares  will 
(by  problem  68)  be  =  «  —  u  —  yY  —  Sz/^  ;  and,  in  the 
very  same  manner,  the  sum  of  the  squares  of  the  two 
terms  {^x  and  2)  adjacent  to  the  middle  one  (tf)  will  be 
=  u^  —  2^/2.       Whence,  by  substituting  these  values, 

our  equations  become  -^  +  u  +  y  •=^  a^  and 

y 

R 


1 22  The  Application  ofAlgebm 


a  —  u  —  yY  —  2z/2  +  iv"  ~2y''  +  y^-by  which,  by  re^ 
duction,  are  changed  to 

aa  ~  2«w  —  2ay  +  2iiu  +  2uy  —  2yy  =  ^, 
and  ay  '^-^    uu  —  ^^  +    2/^  =  ^* 
To  the  former  of  which  add  the  double  of  the  latter ; 

so  shall  aa  —  2au  =  b  -,    and  therefore  w  =  —  —  . 

2  2a 

From  whence,  and  yy  +  a  —  u  X  y  =^  uUj  the  value  of  ^ 
(=  yuu  -I J )  is  likewise  given. 

PROBLEM  LXXII. 

The  sum  («),  the  swn  of  the  squares  (J?)^  and  the  sum  of 
tlie  cub^s  (c),  of  any  four  numbers  in  geometrical  proportion 
being  given  ;  to  find  the  numbers* 

Let  half  the  sum  of  the  two  means  be  x^  and  half  their 
difference  y  ;  also  let  half  the  sum  of  the  two  extremes  be 
2;,  and  half  their  difference  v^  and  then  the  numbers 
themselves  will  be  expressed  thus,  z  —  v^  x  —  z/,  x  +y^ 
z  -^  v:  whence,  by  the  conditions  of  the  problem,  we 
have 

z  —  V  +  X  —  y  -f  X  +  2/  +  z  ^v  =  a^ 
z — vY  +  x—y'Y^x — y^  •\-Z'\-v'Yz:z  b^ 

z  —  V  X  z  -^v  =ix  —  y  X  ^  +  y  (theor.  1,  p.  72)  ;  which, 
contracted,  are 

2z  +2x     z=a^ 
2z^  +  2v^    +  2x^  +  2z/2    =  b^ 
2z^  +  ezv"^  +  2x^  +  6xy^  =  c, 
z^  —     IT  =z    x^  —  y^, 
liCt  x^  —  z^  +  v^j  the  value  of  z/-,  in  the  last  of  these 
equations,  be  substituted  instead  of  2/^,  in  the  two  preced- 
ing ones,  and  wc  shall  have 

2z^  +  2v'^    -f-  2x^  +  2x^  ~  22-    +  2v^    =  b,  and 
22^  +  621^-  +  2x^  +  6x^  —  6xz^  +  6xv^  =  c  ; 
which,  abbreviated,  become 
4x^  +  4v^  =  ^,  and 

Slz^  +  Sx^  —6xz^  +  6x  +  62  X  v^  =  <•. 


to  the  Resolution  of  Problems*  123 

Let  \b  —  AT^,  the  value  of  v^^  in  the  former  of  these 
equations,  be  substituted,  for  its  equal,  in  the  latter^ 
and  we  shall  next  have  2z^  +  ^x^  —  ^xz^  +  6x  +  62  X 
:^f)  —  -2  __.  ^  .  moreover,  if  for  2,  in  the  last  equation, 
its  equal  ^a  —  a:  be  substituted,  there  will  come  out  2  x 
|a  —  xY  +  8^  —  ^x  X  ^CL  —  xf  +  Sa  X  \b  —  XX  z=ici 

that  is,  6ax^  —  Sa^x  -| 1 =  c  ;  therefore  x'^  — 

4  4 

ax         c  b  a^  .  ^,  a 

—  = — *  — ;    and,  consequently,  ,%"  =  —  '— 

2  6a  8  24*  >  1         ^'  4 

—   —  —  +  —  *    whence,    2,   ^,   and  y  are    likewise 
6a         8         48'  ^      '      '  :7 

known, 

T/i^  *awe  otherwise. 


^/ 


Let  the  sum  of  the  two  means  =  ^,  and  their  rect> 
angle  =  r ;    so  shall  the  sum  of  the  two  extremes  =  a 

—  5,  and  their  rectangle  also  =  r  (by  the  questtoii)  : 
from  whence,  and  prob,  68,  it  is  evident,  that  the  sum 
of  the  squares  of  the  means  will  be  =  5^  —  2r,  and 
the  sum  of  the  squares  of  the  extremes  =  a  —  ^j" 
— -  2r;  also,  that  the  sum  of  the  cubes  of  the  means 
will  be  =  s^  — or5,  and  that  of  the  extremes  =  a  —  ^"j* 

—  Zr  X  a  —  5:  by  means  whereof,  and  the  conditions 
of  the  problem,  we  have  given  the  two  following  equa- 
tions  : 

viz*  s^  +  a  —  s\  — •  4r  =  ^,  or,  2*^  —  2as  ^^^r:=^b  —  aa'f 
and  s^  +  a  —  ^1^  —  Zra  =  c,  or  Zas^  —  od^s  —  Zar  =  c  — » 
a?  :  divide  the  former  by  2,  and  the  latter  by  3a,  and  then 

subtract  the  one  from  the  other,  so  shall  r  =  —  -— f- 

6  2 

c  a 

-— ,  whence  the  value  of  5  (= 

3a'  ^      2 


4'- 


-J.  2r  +  — ^   bij  the  first  equation)  is  also 


1 24  The  Application  of  Algebra 

given,    being  (when    substitution    is    made)    =    ~  .— 


I 


aa         b       2c 
12        "i"       3«' 


PROBLEM  LXXIIL 


Having  given  the  sum  («),  and  the  sum  of  the  squared 
(^h)j  of  any  number  of  quantities  in  geometrical  progression; 
to  determine  the  progression* 

Let  the  first  term  be  denoted  by  x^  the  common  ratio 
by  2,  and  the  given  number  of  terms  by  n :  then,  by  the 
conditions  of  the  problem,  we  shall  have 

X  +    xz  +  xz^  +  xz^  +    xz"^  .  .  •  +    xz^^'^^  =  «, 
.v^  +  c^z^  +  xH'^  +  x^z^  +xH^  .  ,  .  +  x^z^"""^  =3  b. 
Multiply  the  first  equation  by  1  — i  2,  and  the  second  by 
1  —  z^  J  so  shall 


.V  —  xz''^    =  a  X  1  —  2;,  and 

Divide  the  latter  of  these  by  the  former ;    whence  will 

be  had  x  +  xz^  =  —  X  1  +  2  :    let  this  equation  and 
a 

the  first  be  now  multiplied  cross-wise,  into  each  other, 
in  order  to  exterminate  x ;     so  shall    a  X    1    +  2"  = 

—  X  1  -f  2  X  1+2+2^+2^   .   .   .   2«-i. 

a 

If  n  be  an  even  number^  put  2w  =  n  \    then  our  last* 
equation,    when    multiplication    by    1   +  2    is    actually 


aa 


made,  will  stand  thus,  ^  x  1  +  2^^"^  =  1  +22+22* 
b 

.   .  .   .    +    2z^'-^    +    222^^-1    +    z^"^',     which,    divid' 

...  aa  \  12 

cd   by  .2-,    becomes  -^+-^^+2-  =  ^  +  ^^;;^::^  + 

~ +  ~+-  +  2+22  +22^  ....  +22^^^ 


to  the  Resolution  of  Problem^  12>5 

+  22>»-i  +  2W»,        Let  s  be  now  put  (s= h  ^)  = 

the  sum  of  the  halves  of  the  two  terms  of  the  series 
adjacent  to  (2)  the  middle  one;  then,  the  rectangle  of 
these  quantities  being  1,  the  sum  of  their  squares  (or 
half  the  sum  of  the  two  terms  of  the  series  next  to 
those)  will  be  =  5^  —  2  (by  problem  68) ;  and  the  sum 

(—  +  2^)  of  half  the' two  next  terms  to  these  last  = 

s^  -—  Zsy  &c.  &c. 

Hence,  by  making  r/  =  —  — ,— ,    and    putting  the 

value  of  —  +  2"*  (as  expressed  in  the  said  problem  68) 

=  Q,  and  then  substituting  above,  &fc.  our  equa- 
tion becomes  dOi  =1+5+5^  —  2+^^  —  3^  + 
^4  —  4^.2  ^  2,  &c.  continued  to  m  terms  ;  whence  the  va- 
lue of  s  may  be  determined. 

Thus,  let  72,  the  number  of  terms  given,  be  four; 
then  m  being  =  2,  Q  (=  ~  +  2^)  will  be  s^  —  2; 
and  our  equation  will,  here,  ht  d  x  6'^  —  2  =  1  -{•  s. 
If  n  be  =  6,  Q  ( =  —   +  2^)   will    be  =  ^^  —  3^  ; 


and  we  shall  have  d  x  s^  —  3,y  =  1  +5+^2  —  2  c= 
52  ^  ^  —  J  .  2ccA  so  in  other  cases,  where  n  is  an  even 
numben 

^  Ifnbe  an  odd  number^  put  2m  =  ?z  ■—  1 ;  and  let  both 
sides  of  the  equation 

c?Xl+2«  =  —  X   1+2  X  1  +  2  +  2^  .  .  .  2»-i 
a 

be  divided  by  1  +  2  ;  so  shall 

a X  1  -2  +2^-2^ . . .  -2"-2^2^-^  =  --  X  1  +  2  +  2^  .  .  +  3«-i 

a 


(because  1  +2X  1  —  2  +  2^  —  2^  +  2'*...--  z'^"^  +  2"-- 


t26  The  Application  of  Algebra 

{1  _  2;  +  2^  —  2^   +  Z"^  ..  .—  2:«-2  ^  2»-i         *  >    _ 

1  ^  2^^  :    whence,  by  transposition,  and  substituting  m, 

«  —  A  X    1   +  22  ^  2^   .   .  .    +  22m  _  a  +  1.   X 
a  a 

2  +  2^    +  z^  .  .  .    2^^! ;  put  22-±J^  =,  c  J  and  let 
the  whole  equation  be  divided  by  a X  2^' ;  then  will 


1 

2^ 

'    ^'^"■^ 

+ 

1 

cx 

2;m-.l  + 

1 

~i    •    • 

+  2'n-3  ^  2*, 


Now,  if  m  be  an  even  number,  the  powers  of  2  in  the 
former  part  of  the  equation  will  be  the  even  ones,  and  those 
in  the  latter  the  odd  ones :  but  if  m  be  an  odd  number, 
then  vice  versa. 

In  the  first  case,  our  equation  may  be  written  thus : 
4  +  "i:i'-  +  -T  +  4-  +  1  +2'  +2'  •  •  •  +  2"^'  +  2-  = 

^-  X  ^  +   4=3  •  •  •  +  4  +  —  +  ^  +  ^'  •  •  •  ^""'  +  2""'  * 

Where,  since 1-  %  =  5,  -^  +  z^  __.  ^^,2  — ^  2,  --   +2^ 

2  2"  2^ 

^  ^.3  —  n^^ j.  2;4  =  54  —  4^2  ^  2^  fe'c.  we  shall,  by  sub- 

2^^ 

stituting  these  values  in  each  series  (proceeding  from  the 
middle  both  ways)  have  1  +  s^TZTi  +  ^^  _  4^^  ^  2  +  £sfc. 
==  c  into  s  +  s^  —  2s  +  fcV. 

But,  in  the  second  case,  where  M  is  an  odd  num- 
ber, and  the  even  powers  of  2  come  into  the  second  series, 

we  shall,  by  the  very  same  method,  have  

s  +  s^  —  3s  +  s^  —  5s^  +  5s  +  ^c.  =  c  into  1  +s^~  2s 
+  ^4  _  4^^2  j^,2  +  ^c. 


to  the  Resolution  of  Problems.  12? 

In  both  which  cases  the  terms  are  be  so  far  conti- 
nued, that  the  exponent  of  5,  in  the  highest  of  them, 

71  —  1 
may  be  =  — > — .      Thus,   if  n^   the  given  number  of 

terms,  be  3,  then  m  {— ~- — )  being  =  1,    the  equa- 

V  tion  belongs  to  case  2,  and  will  be  *  =  c,  barely.  If 
n  =.  5j  then  m  =  2 ;  and  therefore  1  +  s^  —  2  =  C5, 
or  s^  —  1  =  C5,  by  case  1.  If  n  be  7,  m  will  be  3  ; 
and  so  ^  4-  i'^  —  3^  =  c  X  1  +  s^  —  2,  or  s^  —  2^  = 
c  X  s-^  —  1)  ^^  ca^g  2.  Lastly,  if  tz  =  9,  then  vi  =  4, 
and  therefore  1  +s^  —  2  -f-  6-^  --  4a-^  +  2  =  c  Xs  +  s^ — 3^, 
or  s^  —  3s^  +  1  =  c  X  s^  —  2^,  btf  case  1. 

PROBLEM  LXXIV. 

Having  given  the  sum  (a),  and  the  sum  of  the  cubes  (Ji)^ 
of  any  number  of  terms  in  geometrical  progression  ;  to  de^ 
termine  the  progression. 

By  retaining  the  notation  in  the  last  problem,  and  pro- 
ceeding in  the  same  manner,  we  here  have 

a=.x  +  xz  +  xz^  .  .  ,  -L  ATZ"-!  = ^' — ,    and 

2  —  1       ' 

h:=zx''    +  X^Z^  +  X^Z^  .  .  .    +  X^Z^  «-3  _   tz L.    (l^u 

Z^  1        ^  -^ 

theorem  8,  sect.  10). 

Divide  the  last  of  these  equations  by  the  former,  so  shall 

b  _         Z  1X2^^ 1  -         Z^n  4.   2;n    ^    1 

—  =  x^  X  =:x^  X  -z (be- 

a  2^  —  1  X  2'^  —  1  z^  +  z  +  1      ^ 

cause =  2^    I.  2  -f  1,  and  =   z^"-    -f 

2"*  +  1 ).   Let  this  equation,  and  the  square  of  the  first, 

«2  =:  ^2  ^  __ 2! Z — ^  be    now    multiplied,    cross- 

2     —   22    -f-    1 

wise,  in  order  to  exterminate   x;    whence  will  be  had 

^       ^    2^"  _   22"   +    1  ,  2^"    +   2;n    +    1  ,  .   , 

—  X  -T- ^—  =  a^  X      ■       ^■^- :    which, 

-<2  2^      ™   22     +    1  ,  ^2    ^    :^    +    1  ' 


128  The  Application  of  Algebra 

the  numerators  being  divided  by  2",  and  the  denomina- 
tors by  x^  will  stand  thus, 

z'^                                        z^ 
h  X  : =  a^  X -^.        Put  (as 

2;_2+JL  21    +    1     +    2^ 

Z  Z 

before)  the  sum  of  z  and    —  =  5 ;  then,  their  rectan- 
gle being  1,  the  sum  of  their  72th  powers  (z^  +    — ) 

will  be  had  in  terms  of   s  (from  problem  68),    which 
sum  let  be  denoted  by  S  ;  so  shall  our  equation  become 

h  X  — — —  =  a^  X   "^ — '.  whence  the  value  of  s 

s  —  2  s  +  1 

may,  in  any  case,  be  determined. 

Thus  if   (ji)  the  given  number  of  terms  be  3  ;    then 

S  (the  sum  of  the  cubes  of  z  and  — )  being  =5^  —  3*, 

we  have  b  X =  a^  x ;  that 

s  —  2  s  -{-  1 

————————  s^  -       ^o  J.  1 

is,  by  division,  ^  X  6^  +  2^  4-  1  =  a^  X '^^^ 

If  the  number  of  terms  be  4;  then  will  S  =  ^^  — 

45"  +  2  ;  and  therelore  b  x =  ^"^  X : 

s  —  2  5  -f  1 

which,  by  an  actual  division  of  the  numerators,    is    res- 

duced  to  ^  X  'S^  +  2s^  =  a^  X  ^^  —  ^'^  —  36*  -f  3. 

Again,  taking  ii  =  5,  we  have  S  =  6^  —  5s^  -f-  Ss  5 

and  therefore  b  X '■ =  «"*  X 


.9  4-1 


which,  by  division,  is  reduced  to  ^  X  6-^  +  2^^  —  s^  — 2.s  -f  1 
=  a^  X  "S"*  —  s^  —  4^^  -f  4a^  +  1  :  and  so  of  others  ; 
where  it  may  be  observed,  that  the  values  of  S  —  2, 
and  S  +  1,  will  be  always  divisible  by  their  respective 
denominators,  except  the  latter,  when  n  is  either  3,,  or 
a  multiple  of  3. 


io  the  Resolution  of  Problems^  129 

PROBLEM  LXXV. 

The  sum  of  any  rank  of  quantities  (a  +  b  +  c  +  d  + 
e  +  &c.)  being  given  =  P,  the  sum  of  all  their  rectan- 
gles (ab  +  ac  +  ad^  &c.  +  be  -{-  bd^  &c.  +  cd^  &c.) 
=  Q,  the  sum  of  all  their  solids  {abc  +  abd  +  abe^  &c. 
+  acd  +  ace^  &c.  +  bcd^  &c.)  =  R,  &c.  &c.  it  is  pro- 
posed to  determine  the  sum  of  the  squares^  cubeSj  biqiia- 
drateSj  &c.  of  those  quantities. 

rp  =:  b  +  c  +  d^  is?c,  =  sum  of  all  the  quan- 
tities after  the  first  (a), 
,  q  z=.bc  +bd  -\-  be^  &c.  +cd  +  ce^  &c.  =  the  sum 
Put*^  of  their  rectangles, 

r  =  bed  +  bce^  &c.   +  cde^  &c.  =  the  sum  of 
their  solids. 

Then  will  P  =  a    +/^, 

Gi  =  pa  +  q, 

R  =  ya  +  r, 

S  =  r«  +  *, 

T  =.sa  +  t^  &c. 
By  squaring  the  first  of  these  equations,  we  have 
P2  =  (2^  -|-  'jiap  +  p^  y  from  whence  the  double  of  the 
second  being  subtracted  (in  order  to  exterminate  2a/?), 
there  results  P^  —  2Q  =  a^  +  p^  —  2q.  Where 
P^  —  2Q  expresses  the  true  sum  of  all  the  proposed 
squares  (f-  j^  b^  ^  c^  ^  d^  &c. ;  because,  all  the 
quantities  a,  b^  c,  d^  &c.  being  concerned  exactly  alike 
in  the  original,  or  given  equations,  they  must  neces- 
sarily be  alike  concerned  in  the  conclusions  thence  de- 
rived ;  so  that  if  substitution  for  p  and  q  were  to  be  ac- 
tually made  in  the  equation  P^  —  2Q  =.  a^  +  p^  —  2q^ 
here  brought  out,  it  is  evident  that  no  other  dimensions 
of  b^  c,  d^  e^  &c.  besides  the  squares,  can  remain  there- 
in, as  no  dimensions  of  a^  besides  its  square,  have  place  in 
this  equation. 

In  order  to  find  the  sum  of  all  the  cubes, 
put  A(  =  P)=:a-}-^  =  sum  of  the  roots, 
and  B  (  =  P2  _  2Q)  zn  a^  +  p'^  —  2q  =z  sum  of  the. 
squares  ;  then,  by  multiplying  the  two  equations  together, 
we  have  PB  =  a^  +  pa^  yf-  //a  —  2qa  +  //   —  2pq. 


1 30  The  Application  of  Algebra 

From  whence  (to  exterminate  pd^^  the  next  inferior 
power  of  a  after  the  highest,  a^)  let  QA  =  pa^  + 
p^a  +  qa  +  pq  (the  product  of  the  equations  Q  and  A) 
be  subducted ;  and  there  will  remain  PB  —  QA  = 
a^  —  3^<7  +  p^  —  Zpq.  To  this  last  equation  (in  or- 
der to  take  away  the  next  inferior  power  of  a)  add 
three  times  the  equation  R  =  ^a  -f-  r,  so  shall  PB  — 
QA.  +  3R  =  a^  +  p^  —  Spq  +  3r.  From  whence 
it  is  evident  that  PB  —  QA  +  3R  must  be  the  re- 
quired sum  of  all  the  cubes  a\  +  b^  +  c^  +  d^  &c* 
for  reasons  already  specified  with  respect  to  the  preceding 
case. 

To  determine  the  sum  of  the  biquadrates,  put 
C  =  (3^  +  p^  —  Spq  +  3r  =  the  sum  of  all  the  cubes  ; 
tht^n  multiplying  by  the  equation  V  =.  a  +  p  (as  be- 
fore)^ we  get  PC  =  «^  =  pa^  -f-  p'^a  —  opqa  -^  3ra  + 
^4  —  3^2^  ^  r^p^,^  From  which  (to  exterminate  pa^^ 
subtract  QB  =:  pa^  +  p^a  —  2pqa  +  qa^  +  p^q  —  2f 
(the    product    of   the   equations    Q    and    B) :    so    shall 

PC  —  QB  =  «'*  —  qa^  —  pqa  +  3ra  +  p"^  —  4^p^q  + 
Zpr  +  2^2.  tQ  ti^-g  ^^^  j^^  _  ^^2  ^  p^fj^  ^  ^.^  ^  ^^. 

then  will  PC  ~  QB  +  RA  =  «^  +  ^ra  +  p^  —  4/^ 
-f  4pr  +  2q^  :  lastly,  subtract  4S  =  4ra  +  4^,  so  shall 
PC  —  QB  +  RA  —  4S  =  «^  +  /?4  —  4p^q  +  4pr  + 
2^2  —  4^^  _.  j)^  ^j^g  g^j^  q£-  ^y[  the  biquadrates. 

In  like  m.anner  (the  last  equation  being,  again,  mul- 
tiplied by  P  =  «  -f-  /'i  the  preceding  one  by  Q  =  j&a 
-f-  ^,  &c.  &c.)  the  sum  of  the  fifth  powers  will  be 
found  =  PD  —  QC  +  RB  —  SA  +  5T:  from 
whence,  and  the  preceding  cases,  the  law  of  continua- 
tion is  manifest;  the  sum  (F)  of  the  sixth  powers  being 
PE  —  QD  H-  RC  —  SB  -{-  TA  —  6U  ;  and  the  sum 
(G)  of  the  seventh  powers  =  PF  —  QE  +  RD  — 
SC  +  TB  ~  UA  +  rW,  &>.  £ifc. 

But,  if  you  would  have  the  several  values  of  B,  C, 
D,  E,  £sPc.  independent  of  one  another,  in  terms  of  the 
given  quantities  P,  Q,  R,  S,  T,  &fc.  then  will 
B  =  P2  _  2Q, 
C  =  P3  _  3PQ  +  3R, 
D  =  P-*  —  4P2Q  +  4PR  +  2Q2  —  4S, 


to  the  Resolution  of  Problems.  121 

K  =  P^  —  5P3Q  +  5P2R  +  5PQ2  —  5PS  —  5QR 

+  oT,  £s?c.  £?c.  which  values  may  be  continued  on, 
at  pleasure,  by  multiplying  the  last  by  P,  the  last  but 
one  by  —  Q,  the  last  but  two  by  R,  the  last  but  three 
by  —  S,  8Pc.  and  then  adding  all  the  products  toge- 
ther; as  is  evident  from  the  equations  above  derived. — ■ 
These  conclusions  are  of  use  in  finding  the  limits  of 
equations,  and  contain  a  demonstration  of  a  rule,  given 
for  that  purpose,  by  Sir  Isaac  Newton^  in  his  Universal 
Arithmetic. 


SECTION  XIL 


Of  the   Resolution  of  Equations  of  several 
Dimensions. 

BEFORE  we  proceed  to  explain  the  methods  of  re- 
solving cubic,  biquadratic,  and  other  higher  equations,  it 
will  be  requisite,  in  order  to  render  that  subject  more  clear 
and  intelligible,  to  premise  something  concerning  the  ori- 
gin and  composition  of  equations. 

Mr.  Harriot  has  shown  how  equations  are  derived  by 
the  continued  multiplication  of  binomial  factors  into  each 
other;  according  to  which  method,  supposing  X' — a^ 
X  —  b^  X  —  c,  X  —  d^  &c.  to  denote  any  number  of  such 
factors,  the  value  of  x  is  to  be  so  taken,  that  some  one  of 
those  factors  may  be  equal  to  nothing :  then,  if  they  be 
multiplied  continually  together,  their  product  must  also 
be  equal  to  nothing,  that  is,  x  — -a  X  x  —  b  X  ^  — "  c  X 
X  —  <:/,  &c.  =  0  :  in  which  equation  x  may,  it  is  plain,  be 
equal  to  any  one  of  the  quantities,  «,  ^,  c,  d^  &c.  since  any 
one  of  these  being  substituted  instead  of  x^  the  whole  ex- 
pression vanishes.  Hence  it  appears,  that  an'  equation 
may  have  as  many  roots  as  it  has  dimensions,  or  as  are 
expressed  by  the  number  of  the  factors,  whereof  it  is  sup- 
posed to    be   produced.      Thus  the   quadratic  equation 


a  X  X  —  ^  =  0  or  x^ .[   x  -f-  ab  =  0,    has 


1§2  The  Resolution  of  Equations 


two  roots,  a  and  b  ;  the  cubic  equation  x  —  a  X  oc  —  b  X 
a:  —  c  =  0,  or 

—  al  ab'\ 

.^3  _[ z?*  V  a;^  +  (7C  V  ;^  —  abc  =  0,  has  three  roots,  a,  b^ 


and  c  ;  and  the  biquadratic  equation,  x  —  a  X  ^  —  b  X 
X  i —  c  X  ^^  —  d  =0^  or 

-dj  ^^  +  ^'^J  -bed} 

has  four  roots,  a^  b^  c,  and  d.  From  these  equations  it  is 
observable,  that  the  coefficient  of  the  second  term  is  al- 
ways equal  to  the  sum  of  all  the  roots,  with  contrary- 
signs  ;  that  the  coefficient  of  the  third  term  is  always 
equal  to  the  sum  of  their  rectangles,  or  of  all  the  pro- 
ducts that  can  possibly  arise  by  combining  them,  two 
and  two ;  that  the  coefficient  of  the  fourth  is  equal  to 
the  sum  of  all  their  solids,  or  of  all  the  products  which 
can  possibly  arise,  by  combining  them  three  and  three  ; 
and  that  the. last  term  of  all  is  produced  by  multiply- 
ing all  the  roots  continually  together.  And  all  this, 
it  is  evident,  must  equally  hold  good,  when  some  of  the 
roots  are  positive  and  the  rest  negative,  due  regard 
being  had  to  the  signs.        Thus,  in  the  cubic  equation 


:v  —  a  X  oc  —  /9  X  ^  -f  c  =  O,  or  .\^  -I-     —  bYX^  + 
4-  ab" 


+  ab'\ 
-^  ac>  X 
r^bc) 


4-  abc  =R  O  (where  two  of  the  roots,  a,  ^,  are 


positive,  and  the  c^her,  — -  c,  is  negative,  the  coefficient  of 
the  second  term  appears  to  be  —  a  —  b  +  c^  and  that  of  the 
third,  ab  ^ —  ac  —  bc^  or  ab  -i-  a  x  —  c  -}-  b  X  —  c,  con- 
formable to  the  preceding  observations.  .  Hence  it  follows, 
that,  if  one  of  the  roots  of  an  equation  be  given,  the  sum 
of  all  the  rest  will  likewise  be  given ;  and  that,  in  every 
equation  where  the  second  term  is  wanting,  the  sum  of  all 
the  negative  roots  is  exactly  equal  to  that  of  all  the  posi- 
tive ones ;    because,  in  this  case,  they  mutually  destroy 


of  several  Dtmensiom»  133 

each  other.  But  when  the  coefEcient  of  the  second  term 
is  positive,  then  the  negative  roots,  Jaken  together,  ex- 
ceed the  positive  ones.  But  the  negative  roots,  in  any 
equation,  may  be  jchahged  to  positive  ones,  and  the  posi- 
tive to  negative,  by  changing  the  signs  of  the  second,  fourth, 
and  sixth  terms,  tol  iso  on,  alternately.  Thus,  the  fore- 
going equation 

_        +ab^^ 

{x  —  a  X  x^^b  X  ^  +  (?  =  )  x^  + 


—  al        +  ab't 

—  l?>  x^  —^  ac>  X  - 
+  c}       —  be} 


abc  =  Oj  by  changing  the  signs  of  the  second  and  fourth 

+  «1  +  «^1 

terms,  becomes  x^  +  +bYx'^  —  ac>x  —  abc  =  0,  or 

—  ^J         —bcS 

X  -^-a  X  oc  +  b  X  ■^'  — <-'  ==  0;  where  the  roots,  from 
-f-  «,  +  6,  and  ■;-—  c,  are  now  become  —  «,  —  ^,  and  +  c*. 
Moreover,  the  negative  roots  may  be  changed  to  positive 
ones,  or  the  positive  to  negative,  by  increasing  or  di- 
minishing each,  by  some  known  quantity.  Thus,  in  the 
quadratic  equation  x^  +  8x  +  15  =  0,  where  the  two 
roots  are  —  3  and  —  5  (and  therefore  both  negative) 
if  z  —  7  be  substituted  for  x^  or,  which  is  the  same  thing, 
if  each  of  the  roots  be  increased  by  7,  the  equ^ion  will 
become  z  —  7^  -*-  8X2  —  r  -f  15  =  0  ;  that  is,  z^  — 
6z4-8=:0,  orz  —  2  X  2  —  4  =  0;  where  the  roots 
are  2  and  4,  and  therefore  both  positive.  This  method 
of  augmenting,  or  diminishing  the  roots  of  an  equation 
is  sometimes  of  use  in  preparing  it  for  a  solution  by 
taking  away  its  second  term  ;  which  is  always  perform- 
ed by  adding  or  subtracting  ^,  A^  or  |.  part,  qs?c.  of  the 
coefficient  of  the  said  term,  according  as  the  proposed 
equation  rises  to  two,  three,  or  four,  fcPc.  dimensions. 
Thus,  in  th^  quadratic  equation  x^  — •  Sx  -f-  15  =  0,  let 
the  roots  be  diminished  by  4,  that  is,  let  x  —  4  be  put 
=  2,  or  .V  =  4  -f  2  :  then,  this  value  being  substituted 
for  X,  the  equation  will  become  z  +  4]^  - — S  X  z  -^-4  + 
15  =  0,  or  ;o^  —  1=0;  in  which  the  second  term  is 
wanting. 


1 34  The  Resolution  of  Equations 

Likewise,  the  cubic  equation  2'  -^  az^  +  bz  —  c  =0, 
by  writing  a;  = 4-  ^^  and  proceeding  as  abovis:, 

o 

will  become  x^^       1  2r^  +  —     ^     ^=0;  and  so  of 

Others. 

Hence  it  appears,  how  any  adfected  quadratic  may  be 
reduced  to  a  simple  quadratic,  and  so  resolved  without 
completing  the  square  ;  but  this  by  the  bye.  I  now  pro- 
ceed to  the  matter  proposed,  viz,  the  resolution  of  cubic, 
biquadratic,  and  other  higher  equations ;  and  shall  begin 
with  showing 

Ilorv  to  determine  -whether  some^  or  all  the  roots  of  an  equa^ 
tio?i  be  rational^  and^  f^^j  "i^hat  they  are. 

Find  all  the  divisors  of  the  last  term,  and  let  them  be 
substituted,  one  by  one,  for  x  in  the  given  equation  ;  and 
then,  if  the  positive  and  negative  terms  destroy  each  other, 
the  divisor  so  substituted  is  manifestly  a  root  of  the  equa- 
tion ;  but  if  none  of  the  divisors  succeed,  then  the  roots, 
for  the  general  part,  are  either  irrational  or  impossible  : 
for  the  last  term,  as  is  shown  above,  being  always  a  multi- 
ple of  all  tfie  roots,' those  roots,  when  ratioiial,  must  neces- 
sarily be  in  the  number  of  its  divisors. 

Examp.  1.   Let  the  equation  x^  — 4:^;^  —  7x  +  \0  =zO^ 
be  proposed  ;  then  the  divisors  of  (10)  the  last  term  being 
+  1^  —  1^  ^  2,  —  2,  -f.  5,  —  5,  +  10,  —  10,  let  these  ^ 
quantities  be  successively  substituted  instead  of  ;c',  and  we 
shall  have, 

1  —      4  — ,    7-f-10=    0,  therefore  1  is  a  root ; 

—  1  —      4  -|»    7  +  10  =  12,  therefore  —  1  is  no  root ; 
8  —    16  —  14  -f  10  =  —  12,  therefore  2  is  no  root  ,• 

—  8  —    16  -f.  14  -f  10  =  O,  therefore  — 2  is  anodier  root ; 
125  —  100  —  35  -f-  10  =  O,  therefore  5  is  the  third  root. 

It  sometimes  happens  that  the  divisors  of  the  last 
term  are  very  numerous  ;  in  which  case,  to  avoid  trou- 
ble, it  will  be  convenient  to  transform  the  equation  to 
another,  wherein  the  divisors  ave  fewer  ;  and  this  is  best 


of  several  Dimensions.  ,135 

effected  by  increasing  or  diminishing  the  roots  by  a  unit, 
or  some  other  known  quantity. 

Examp*  2.  Let  the  equation  propounded  be  y^  —  4z/^  — . 
8z/  +  32  =  O;  and,  in  order  to  change  it  to  another, 
whose  last  term  admits  of  fewer  divisors,  let  ;i^  +  1  be 
substituted  therein  for  y^  and  it  will  become  oc"^  —  Gat^  — 
♦  16a:  +  21  =  0;  where  the  divisors  of  the  last  term  are 
1,  —  1,3,  —  3,  7,  —  7,  21,  and  — ^  21 ;  which  being  suc- 
cessively substituted  for  x^  as  before,  we  have, 

1  —    6 — 16  +  21=0,  therefore  1  is  one  of  the  roots  ; 

1  —  6  +  16  +  21  =  32,  therefore  —  1  is  not  a  root ; 
81  —  54  — 48  +  21  =  0,  therefore  3  is  another  root. 
But  the  other  two  roots,  without  proceeding  further, 
will  appear  to  be  impossible  ;  for,  their  sum  being  equal 
to  - —  4,  the  sum  of  the  two  positive  roots  (already  found), 
-with  a  contrary  sign  (as  the  second  term  of  the  equation 
is  here  wanting),  their  product,  therefore,  cannot  be 
equal  to  (7)  the  last  term  divided  by  the  product  of  the 
other  roots,  as  it  would,  if  all  the  roots  w^ere  possible. 
However,  to  get  an  expression  for  these  imaginary  roots, 
let  either  of  them  be  denoted  by  t;,  and  the  other 
will  be  denoted  by  —  4  —  Vy  which,  multiplied  toge- 
ther, give  —  4^v  —  ^-  =  7  ;  whence  v  =  —  2  +  V —  3, 
and  consequently  — 4  —  v  =  —  2-—  V  —  3.  Now 
let  each  of  the  four  roots,  found  above,  be  increased  by 
unity,  and  you  will  have  all  the  roots  of  the  equation 
proposed. 

When  the  equation  given  is  a  literal  oiiCj  you  may 
still  proceed  in  the  same  manner,  neglecting  the  known 
quantity  and  its  powers,  till  you  find  what  divisors  suc- 
eeed ;  for  each  of  these,  multiplied  by  the  said  quantity, 
will  be  a  root  of  the  equation.  Thus,  in  the  literal  equa- 
tion, x^  +  3rt;c^  —  4<a^x  —  12a^  =  O,  the  numeral  divisons 
of  the  last  term  being  1, . —  1,  2,  —  2,  3,  —  3,  i^c.  I  write 
these  quantities,  one  by  one,  instead  of  x^  not  regarding. 
a ;  and  so  have 

1+3  — 4  —  12  =  —  12,  therefore  a  is  not  a  root ; 

' —  1-f    3+4  —  12  =  - —    6,  therefore  —  «  is  no  root ; 

8  +  12  — '8-^—12  =  0,  therefore  2«  is  one  of  the  roots ; 


136  The  Resolution  of  Equations 

—  8  +  12+  8  —  12  zrO,  therefore — 2a  is  another  root ; 
27  +  27 — 12  —  12  =  30,  therefore  3«  is  not  a  root; 
_  27  +  27  + 12  —  12  =0,  therefore  —3a  is  the  third  root. 
The  reason  of  these  operations  is  too  obvious  to  need  a 
further  explanation.  I  shall  here  subjoin  a  different  way, 
whereby  the  same  conclusions  may  be  derived,  from  Sii* 
Isaac  Newton''s  Method  of  Divisors  ;  which  is  thus  : 

Instead  of  the  unknown  quantity^  substitute^  successively^ 
three  or  more  adjacent  terms  of  the  arithmetical  progres- 
sion^ 2,  1,  0,  —  1,  —  2  ;  and^  having  collected  all  the 
terms  of  the  equation  into  one  sum^  let  the  quantities  thus 
resulting^  together  with  all  their  divisors^  be  placed  in  a 
line^  right  against  the  corresponding  terms  of  the  progres- 
sion 2,  1,0,  —  1,  —  2  ;  then  seek  among  the  divisors  an 
arithmetical  progression^  whose  terms  correspond  with^  or 
stand  according  to  the  order  of  the  terms  2, 1,  0,  —  1,  — -  2, 
of  the  first  progression^  and  whose  common  difference  is 
either  a  iinit^  or  some  divisor  of  the  coeffcie7it  of  the  highest 
power  of  the  unknozvn  quantity  {x)  in  the  given  equation. 
If  any  such  progression  can  be  discovered^  let  that  term  of  it 
which  stands  against  the  term  0,  in  the  first  progression^  be 
divided  by  the  common  difference^  and  let  the  quotient^  xvith 
the  sign  +  or  — prefixed^  according  as  the  progression  is 
increasing  or  decreasing^  be  triedy  as  above^  by  substituting 
it  for  X  in  the  proposed  equation. 

Thus,  let  the  proposed  equation  be  x^  —  a^  —  lOx 
+  6  =  0;  then,  by  substituting  successively  the  terms  of 
the  progression  2,  1,  0,  —  1,  instead  of  at,  there  will  arise 
—  10,  —  4,  6,  and  14,  respectively  ;  which,  together  with 
their  divisors,  being  placed  right  against  the  correspond- 
ing terms  of  the  progression  2,  1,  0,  —  1,  the  work  will 
stand  thus : 


2 

—  10 

1.2.5. 

10 

5 

1 

—   4 

1.2.4. 

4 

0 

+     6 

1.2.3. 

6 

3# 

1 

+  14 

1.2.7. 

14 

2 

Now,  since  the  coefficient  of  the  highest  power  (a:^) 

•is  here  only  divisible  by  a  unit,  I  seek,  among  the  di- 

t^isors,  a  collateral  progression  whose  common  difference 


ijf  several  Dimensiom*. 


1^7 


is  a  unit ;  and  find  the  only  one  of  this  kind  to  be  5^  4,  3, 
2 ;  whose  third  term  standing  against  the  term  0  in  the 
first  progression,  I  therefore  take  and  divide  by  unity,  and 
then  substitute  the  quotient,  with  a  negative  sign,  instead 
of  x^  and  there  results  —  27  —  9-f-30  +  6  =  0;  there* 
fore  —  3  is,  manifestly,  a  root  of  the  equation. 

Again,  if  the  proposed  equation  were  to  be  2x^  —  5x^ 
-f  4,x  —  10  =  0,  we  shall,  by  proceeding  in  the  same  man- 
ner, have 


2 

—    6 

1 

—    9 

0 

—  10 

—  1 

—  21 

—  2 

—  54 

In   which 

case,   '. 

1.2.3.     6 

1 

1.3.9 

3 

1   .  2  .  5  .   10 

5# 

1   .  3  .   7  .  21 

7 

1.2.3.     6  . 

9&C. 

9 

]  discover,  among  the  divisors,  the 
increasing  arithmetical  progression,  1,  3,  5,  7,  9  ;  whose 
third  term,  5,  standing  against  the  term  0  in  the  first 
progression,  being  divided  by  2,  the  common  difference, 
and  the  quotient  (|)  substituted  for  x^  the  business  suc- 
ceeds, the  positive  and  negative  terms  destroying  each 
other. 

Moreover,  if  the  equation  x^  +  ^^  —  29x^  — *  9:^  +  180 
=  0  were  proposed,  the  work  will  stand  as  follows  : 


2 

70 

1 

144 

0 

180 

—  1 

160 

—  2 

90 

1  .2.  5  .  7.  10.  14.  35  .  70 

1 

2 

5 

7 

1.2.3.4.    6  .    8  .    9  .  12  &c. 

2 

3 

4 

6 

1.2.3.4.    5  .    6  .    9  .  10  &c. 

3 

4 

3 

5# 

1  .2  .4.  5  .    8  .  10.  16.  20  &c. 

4 

5 

2 

4 

1.2.3.5.    6.    9.  10.  15  &c. 

5 

6 

1 

3 

discovered  no  less  than  four  pi 

'Og 

res 

>sic 

)ns, 

whose  terms  differ  by  unity ;  whereof  the  terms  corre- 
sponding to  the  term  0,  in  the  first  progression,  are  3,  4, 
3,  and  5  :  therefore  the  two  former  progressions  being 
ascending  ones,  and  the  two  latter  descending,  I  try  the 
quantities  +  3,  +  4,  —  3,  —  5,  one  by  one,  and  find  that 
they  all  succeed. 

And  after  the  same  manner  we  may  proceed  in  other 
cases;  but,  in  order  to  try  whether  any  quantity  thus 
found  be  a  true  root,  we  may,  instead  of  substituting 
for  .V,  divide  the  whole  equation  by  that  quantity  coi>* 


i  58  The  Resolutmi  of  Equations^ 

nected  to  x^  with  a  contrary  sign  ;  for,  if  the  division  ter- 
minate without  a  remainder,  the  said  quantity  is,  mani- 
festly, a  root  of  the  equation. 

Thus,  in  the  last  example^  where  the  equation  is 
x^  +  x^  —  29x^  —  9a:  +  180  =  O,  the  numbers  to  be 
tried  being  +  3,  +  4,  —  3,  and  —  5,  I  first  take  —  3 
and  join  it  to  at,  and  then  divide  the  whole  equation,  x"^ 
+  x^  —  29x^  —  9x  +  180  (=  0),  by  a:  —  3,  the  quantity 
thence  arising,  and  find  the  quotient  to  come  out  x^  + 
4a:^  —  17 x  —  60,  exactly.  Therefore  +  3  is  on€  of  the 
roots. 

Again,  in  order  to  try  +  4,  the  second  number^  I  divide 
the  quotient  thus  found,  by  a;  —  4,  and  there  comes  out 
x^  +  Sx  +15;  therefore  +  4  is  another  root :  lastly,  I 
try  —  3,  by  dividing  the  last  quotient  by  ■%■  +  3,  and  find 
It  also  to  succeed,  the  quotient  being  x  +  5*  See  the  ope- 
ration at  large. 

,v„3)  x^  +   x^  —  29;^^— 9:v  +  180(;c^+4Ar*  —  17x — 60 

x"^! — Sx^ 

^4x^ — 29a;2 
4-  4x^.^  12;c^ 


— 17;%2--   9x 
^17x^  -hSlx 

—  eOx  -f  180  ' 

—  60x  +  180 


O  O 


.  4)  .v^  +  4x^  —  17x  —  60  (a:^  +  Sat  +  15 


x^>^4x^ 

+  8^:^—  17x 
4-  Sx^  —  32a;' 

+  15a:- 
-4-15A:- 

-60 
-60 

0 

o 

gJ several  Dimensions*  139 

X  +  S)  x^  +  8x  -{-  15  {x  +  5 
x^  +  Sx 


+  5X  + 
^5x^ 

15 
15 

0 

0 

As  another  instance  hereof,  let  there  be  proposed  the 
equation  2x^  ~  3;^^  +  16:\:  > —  24  =  0  ;  then,  expounding 
X  by  2,  1,  0,  and  — •  1,  successively,  and  proceeding  as  in 
the  foregoing  examples,  we  have 


2 

+  12 

1.2.3.4. 

6  .   12 

+  2 

—  1 

1 

—    9 

1.3.9 

+  3 

+  1 

0 

—  24 

1.2.3.4. 

6  .     8  &c. 

+  4 

+  3^ 

1 

—  45 

1,3,5.9. 

15  .  45 

+  5 

+  5 

Therefore,  the  quantities  to  be  tried  being  4  and  |, 
I  first  attempt  the  division  by  at  •—  4  ;  which  does  not 
answer :  but,  trying  -x*  —  ^,  or  (its  double)  2:v  — -  3, 
I  find  it  to  succeed,  the  quotient  being  x^  +  8,  ex- 
actly. 

The  reason  why  the  divisors,  thus  found,  do  not  always 
succeed,  is,  because  the  first  progression,  2,  1 ,  0,  -—  1  is 
not  continued  far  enough  to  know  whether  the  corres- 
ponding progression  may  pot  break  off,  after  a  certain 
number  of  terms  ;  which  it  neyei:  can  do  when  the  bu- 
siness succeeds.  Thus,  in  the  last  example,  where  we 
had  two  different  progressions  resulting^  had  the  operation, 
or  series,  2,  1,  0,  —  1,  been  continued  only  two  terms 
farther,  you  would  have  found  the  first  of  those  pro- 
gressions to  fail ;  whereas,  on  the  contrary,  the  last  (by 
which  the  business  succeeds)  will  hold,  carry  on  the  pro- 
gression, 2,  1,  0,  -r- 1,  as  far  as  you  will.  The  grounds  of 
which,  as  well  as  of  the  whole  method,  upon  which  thq 
foregoing  observations  are  founded,  may  be  explained  in 
the  following  manner. 

Let,  there  be  assumed  any  equation,  as  ax^  +  bx^  -f- 
cx^  +  dx  -{^  e  =z  %  wherein  «,  ^,  c,  d^  and  ^,  represent 
any  whole  numbers,  positive  or  negative,  and  let  px  +  q 
denote  any  binomial  divisor  by  which  the  said  expression 


140  The  Resolution  of  Equations 

ax^  +  bx^  +  cx^  +  dx  +  e  \^  divisible,  and  let  the  quo- 
tient thence  arising  be  represented  by  rx^  -f-  sx^  ^  tx  -{-  v, 
or,  which  is  the  same  in  effect,  let  ax^  -f-  bx^  +  cx'^  + 
dx  -{-  e  ^^  px  -\-  q  X  r..^  -f  sx^  -f-  tx  -^  v.  This  being 
premised,  suppose  x  to  be  now,  successively,  expounded 
by  the  terms  of  the  arithmetical  progression  2,  1,  0,  '^—  1, 
«— 2  {as  abcvey-,  and  then  the  corresponding  values  of 
our  divisor  px  +  q^  will,  it  is  manifest,  be  expounded  by 
2p  -f-  q^  p  +  q^  q^  — P  +  9f  ^^^  ' —  ^P  +  ^i  respectively  ; 
which  also  constitute  an  arithmetical  progression,  whose 
common  difference  is/?;  which  common  difference  (/?) 
must  be  some  divisor  of  the  coefficient  (a)  of  the  first 
term,  otherwise  the  division  could  not  succeed,  that  is,  p 
could  not  be  had  in  a  without  a  remainder.. 

Hence  it  appears  that  the  binomial  divisor,  by  which  an 
expression  of  several  dimensions  is  divisible,  must  always 
vary  as  x  varies,  so  as  to  be,  successively,  expressed  by  the 
terms  of  an  arithmetical  progression,  whose  common  dif- 
ference is  some  divisor  of  the  first,  or  highest  term  of  that 
expression. 

It  also  appears  that  the  said  common  difference  is  al- 
ways the  coefficient  of  the  first  term  of  the  general  divi- 
sor ;  and  that  the  term  {q)  of  the  progression,  which  arises 
by  taking  :v  =  0,  is  the  second  term.  Therefore,  when- 
ever, by  proceeding  according  to  the  method  aboye  pre- 
scribed, a  progression  is  found,  answering  to  the  conditions 
here  specified,  the  terms  of  that  progression  are  to  be  con- 
sidered only  as  so  many  successive  values  of  some  general 
divisor,  as  px  -f-  q.  Whence  the  reason  of  the  whole  pro- 
cess is  manifest. 

After  the  same  manner  we  may  proceed  to  the  in- 
vention of  trinomial  divisors,  or  divisors  of  two  dimen- 
sions ;  for,  let  mx^  +  px  +  q,  be  any  quantity  of  this 
kind,  wherein  772,  /?,  and  q  represent  whole  numbers 
positive  or  negative,  and  let  the  terms  of  the  progres- 
sion 3,  2,  1,  0,  . —  1,  - — '2,  ~  3,  be  written  therein,  one 
by  one,  instead  of  x ;  whence  it  will  become  9m  -f  3p 
^  q^  4m  +  2p  +  q,  m  +  p  +  q,  q,  m  ^p  +  q,  4m  — 
,  2p  4-  ^,  and  9m  —  Sp  +  ^,  respectively  ;  where  m  must 
be  ^ome  divisor  of  the  coefficient  of  the  first  term  of  the 


of  several  Dimensions^  141 

given  expression  ;  otherwise,  the  division  could  not  suc- 
ceed.    Hence  it  appears, 

1°,  That  the  coefficient  (jn)  of  the  first  term  of  the 
divisor  must  always  be  some  numeral  divisor  of  the  co- 
efficient of  the  first  term  of  the  proposed  expression, 

2°,  That  the  product  of  that  coefficient  by  the  square 
of  each  of  the  terms  of  the  assumed  progression,  3,  2,  1, 
Oj  —  1,  —- *  2,  •—  3,  being  subtracted  from  the  corre- 
sponding value  of  the  general  divisor,  the  remainders 
{Zp  +  q,  2p  +  q,  p+  q,  q,  —p  +  q,  ~2p  +  q,  ~  Sp 
-f-  q)  will  be  a  series  of  quantities  in  arithmetical  pro- 
gression, whose  common  difference  is  the  coefficient  of  the 
second  term  of  the  divisor. 

3°.  And  that  the  term  (^)  of  this  progression,  which 
arises  by  taking  ^t-  =  0,  will  always  be  the  third,  or  last 
term  of  the  said  divisor.  From  whence  we  have  the 
following  rule.  Instead  of  x  in  the  quantity  proposed^ 
substitute^  successivelij^  four  or  more  adjacent  ter?ns  of  the 
progression  3,  2,  1,  0,  — -  1,  —  2,  — -  3  ;  and  from  all 
the  several  divisors  of  each  of  the  numbers  thus  residting^ 
subtract  the  squares  of  the  corresponding  terms  of  that  pro- 
gression multiplied  by  some  numeral  divisor  of  the  highest 
term  of  the  quantity  proposed^  and  set  doxvn  the  remainders 
right  against  the  corresponding  terms  of  the  progression  3, 
2,  1,  0,  —  1,  —  2,  —  3;  and  then  seek  out  a  colla- 
teral progression  which  runs-  through  these  remainders ; 
which  being found^  let  a  trinornial  be  assumed^  whereof  the 
coefficie7it  of  the  first  term  is  the  foresaid  numeral  divisor  ; 
that  of  the  second  term^  the  common  difference  of  this  col- 
lateral progression  ;  and  xvhereof  the  third  term  is  equal  to 
that  term  of  the  said  progression  which  arises  by  taking 
Oi  -=.0  ;  and  the  expression  so  assumed  will  be  the  divisor  to 
be  tried.  But  it  is  to  be  observed^  that  the  second  term 
raust  have  a  negative  or  positive  sign^  according  as  the  pro- 
gression^ found  among  the  divisors^  is  an  increasing  or  a 
decreasing  one. 

Thus,  let  the  quantity  proposed  be  x'*'  —  x'^  —  5x^  -f- 
12^  —  6;  and  then,  by  substituting  3,  2,  1,  0,  —  1, 
—  2,  successively,  instead  of  x,  the  numbers  resulting 
will  be  39,  6,  1,  —  6,  —  21,  and  —  26,  respectively; 
which,  together  with  all  their  divisors,  both  positive  and 


142 


The  Resolution  of  Equations 


negative,  I  place  right  against  the  corresponding  terms  of 
the  progression  3,  2,  1,  0,  - —  1,  ' — •  2,  in  the  following 
manner : 


1 
0 

—  1 

—  2 


39, 
6. 
1  .' 
6. 

21  . 

26. 


13.3 


-  1 

3.2 

7.S 

13.2 


—  1 

—  1 

—  1 


—  2  .  —     3  • 

—  3  .  ~     7, 
, ,  2  .  13 


39 
6 

6 
21 
26 


Then,  from  each  of  these  divisors  I  subtract  the  square 
of  the  corresponding  term  of  the  first  progression  muhi- 
plied  by  unity  (as  being  the  only  numeral  divisor  of  the 
first  term),  and  the  work  stands  thus  : 


3 

30.     4.-6—8 10.^12.-22—48 

+  4 

—  6 

2 

2._1.^2 3.~  5.—  6 7.— 10 

+  2 

—  3 

1 

0 

+  0 
2 

+  0 

6.     3.     2.      1. —  1.—  2. —  3 — .  6 

-1 

20.     6.     2.     0.— T  2.—  4 8.-22 

—  4 

+  6 

-2 

22.     9.-2 — 3 r  5 -  6.— 17 — 30 

—  6 

+  9 

Here  I  discover,  among  the  remainders,  two  colla- 
teral progressions,  viz.   4,  2,  0,  —  2,  —  4,  —^  6,  and 

—  6,  —  3,  0,  -f-  3,  +  6,  -f  9  ;  therefore  the  quantity 
to  be  tried  is  either  x^  +  2x  —  2,  or  Ai-^  —  Sx  +  3  ;  by 
both  of  which  the  business  succeeds. 

This  invention  of  trinomial  divisors  is  sometimes  of 
use  in  finding  out  the  roots  of  an  equation  when  they 
are  irrational,  or  imaginary.  Thus,  let  the  equation 
given  be  x"^  —  4^^'  -f  5x^  —  4.v  —  1=0;  and  let  x  be 
successively  expounded  by  the  terms  of  the  progression 
3,  2,  1,  0,  and  the  numbers  resulting  will  be  7,  —  3, 

—  1,  and  1  ;  which,  together  with  their  divisors,  being 
ordered  according  to  the  preceding  directions,  the  ope-r 
ration  will  stand  as  follows  : 


5 

7  .     1 

— 1  .—7 

— 2  .—8  .—10—16 

2 

3  .     1 

.— 1 3 

_1  ._3  ._  5—  7 

1 

1  .—1 

#         # 

0  —2          ^       ^^ 

0 

1   .—1 

#         # 

1  —1          ^        ^ 

,2 

— 1 

0 

+  1 


— 5 

— 2 
+  1# 


Here  we  have  two  progressions,  —  2,  —  1,  0,  1 ;  and 
—  8,  —  5,  —  2,  1;  therefore  the  quantity  to  be  tried 
is  either  x^  —  ;c  +  1,  or  ^^  —  ?^x  -f  1  ;  but  I  take  the 


tf  several  Dimensions.  143. 

first,  and  having  divided  x"^  ^—  4x^  +  5x^  —  4x  +  1 
thereby,  find  it  to  succeed,  the  quotient  coming  out 
x^.^3x  +  1,  exactly.  Therefore  x"^  —  4x^  +  5x^  — 
4;v  +  1  being  universally  equal  to  x^  —  x  +  1  X 
;^2  —  Sx  +  1,  let  ;<:2  —  X  +  1  be  taken  =  0,  and  also 
x^  —  3;c  -f-  1  =  0 ;  from  the  former  of  which  equations 
we  have  x  =z  |  ±V  —  | ;  and  from  the  latter  x  =:  ^  ± 
VK^  Therefore  the  four  roots  of  the  given  equation 
^rU+Vtrj^  1  _  V""::^^,  I  +  V^andf  — V|l 
whereof  the  two  last  are  irrational  and  the  two  first 
imaginary.  And  in  the  same  manner,  the  roots  of  a  li- 
teral equation,  as  z^  —  4az^  +  5a^z^  —  4a^z  +  a^  =  0, 
where  the  terms  are  homogeneous,  may  be  derived ;  for, 
let  the  roots  be  divided  by  «,  that  is,  let  x  be  put  = 

^  ^  or  ax  =  z  ;  and  then,  this  value  being  substituted  for 

a 

z,  the  equation  will  become  x"^  -—  4^;^  +  5x^  —  4^  +  1 

==  0 ;  from  which  x  will  be  found,  as  above  j  whence  z 

(  =  ax^  is  also  known. 

Having  treated  largely  of  the  manner  of  managing 
such  equations  as  can  be  resolved  into  rational  factors, 
whether  binomials  or  trinomials,  I  come  now  to  ex- 
plain the  more  general  methods,  by  which  the  roots  of 
equations,  of  several  dimensions,  are  determined ;  and 
shall  begin  with 

The  Resolution  of  cubic  Equations^  according  to  Cardan. 

If  the  given  equation  have  all  its  terms,  the  second 
term  must  be  taken  away,  as  has  been  taught  at  the  be- 
ginning of  this  section;  and  then  the  equation  will  be 
reduced  to  this  form  :  viz.  x^  +  ax  z=:  b  \  where  a  and  b 
represent  given  quantities.  Put  x  =:  y  +z  ;  and  then, 
this  value  being  substituted  for  x^  our  equation  becomes 
i/  +  3t/^2jf%2;2  ^  z^  +a  X  y  '+  z  =  ^,  or  y^  +  z^ 
+  Syz  X  y  -i-z+axy+z  =  b.  Assume,  now,  3yz 
=  —  a ;  so  shall  the  terms  Syz  X  y  +z  and  a  X  y  +  z 
destroy  each  other,  and  our  equation  will  be  reduced  to 
y^  +Z^  =:  b.  From  the  square  of  which,  let  four  times 
the  cube  of  the  equation  yz  =z  —  ^a  be  subtracted,  and 


144  The  Resolution  of  Equations 

we  shall  have  t/«  —  Si/^z^  4.  :^«  -=  Z)«  4.  __  ;   and  there- 
fore, by  extracting  the  square  root,  on  both  sides,  y^  — 

"i? 

from  2/3  +  2^  =  ^,  gives  2t/^  =  ^  +  y  Z^^  -f-  - — ,  and 


/  46/^ 

=  \^^  +  -^  ;     which    added    to,    and    subtracted 


2z3  =  b  -J^±t:  hence  y^l  +  JH  +  t} 

;    and  consequently  x  (^z=z  y 


b  jo^       a^ 

and  2  =  —  « —  V 1 

2         ^  4  ^  27 


+  -)  =  T  +  V4  +27I    -^T-slj  +  TrV 

which  is  Cardaii!^  theorem :  but  the  same  thing  may 
be  exhibited  in  a  manner  rather  more  commodious  for 
practice  by    substituting  for  the  second  term  its  equal 

—  \^     

b  1 6^        a^l'^  (=   — -^  =  2;,  because  uz  =z  — 

T  +  VT+i?!  ^ 

|a).      And,  this  being  done,  our  theorem  stands  thus  : 


"^^^4     '27! 


b        lo- 
—  +V h 

2     '   >  4    ^' 


Example  1.  Let  the  equation  ^'  +  Sz/^  +  Qy  =  13  be 
propounded ;  and,  in  order  to  destroy  the  second  term 
thereof,  let  ;v  .—  1  be  put  :=s:  y;  so  shall  x  ' —  Ip  + 
3  X  ^' —  iT  +  9  X  ^^  —  1  =  13,  or  x'^  +  6x  =  20 ; 
therefore,  in  this  case,  a  being  =  6,  and  b  =  20,  we 

j  +  ^^+Tri 


of  several  Dimensions^  145 

10^+  VlooTsl^  —  ^  -TY  =  20,3923]^ 

10  + V 100 +  8  b 

2 
- —  =  2.732  —  .732  =  2  ;    and  consequently  y 

20,3923]^ 

(  =  :^— .1)=:1. 

Exam.  2.  If  the  equation  given  be  y^  —  32/^*  — •  2y'^ 
. —  8  =  0;  then,  by  writing  x  +  1  for  z/^,  it  will  be- 
come T+T]'  —  3  X  ^  +  iT  —  2X^  +  1  —  8  =  0, 
or'  ::c'  —  5x  ■=•    12  :     therefore,  ^  being  =  —  5,  and 

^  =  12,  .V  will  here  be  equal  to  6  +  \/36  —  ^i^\^  — 

—  4 1 1  1 ,6666,  y c. 

—  -^====^  =  6  +  5,6009>  +  — ^ 

6  +  \/36  —  VV  l"^  ^   +  5,6009  1 3 

=  2,26376  +  ,73624  =  3 ;  and  consequently  y^ 
(=.v  +  l)=4:  which  is  the  only  possible  value  of 
y^  in  the  given  equation.  And  it  will  be  proper  to  take 
notice  here,  that  this  method  is  only  of  use  in  cases 
where  two  of  the  three  roots  are  impossible  (except  when 

b^        a^       .        , 
they  are  equal)  ;    for 1 being,  in  all  other  cases, 

a  negative  quantity,  its  square  root  is  manifestly  impos- 
sible. 

I  shall  now  give  the  investigation  of  the  same  ge- 
neral theorem,  for  the  solution  of  cubics,  by  a  different 
method ;  which  is  also  applicable  to  other  higher  equa- 
tions. 

Supposing,  then,  the  sum  of  two  numbers,  z  and  z/, 
to  be  denoted  by  5,  and  their  product  (zy^  by  /?,  it 
will  appear  (^from  prob.  68,  p*  119)  that  the  sum  of 
their  cubes  (z^  +  z/^)  will  be  truly  expressed  by  6*^  — 
3j&5. 

If,  therefore,  z^  +  y^  be  assumed  =  ^,  we  shall  also 

have  s^  —  ops  =  b  :    but,  zy  being  =  /?,  or  y  =  — ,  our 


z 

/i3 


lirst  equation  z^  '\-  if  =.  b^  will  become  z^  +  ^  =  Z' ; 

from  which,  by  completing  the    square,  &Pc.  z  is  found 

U 


146  I' fie  Resolution  of  Equations 

']  z=.\h  '\-  \^\ob  —  p^\^\    whence  j/  (  =  ~)  is  given  = 

P 
— ~ ~  ;    and  consequently  «  (  =  2;  +  J/) 


=  i|^  ^\\t)h — /^Js   ^  _P_ —  .    whicl>  is, 

evidently,  the  true  root  of  the  equation  6^  < —  '^ps  =  h* 
From  whence  the  root  of  the  equation  o<^  +  ax  =  b^ 
wherein  the  second  term  is  positive,  will  be  given,  by 
writing  x  for  .9,  and  ^a  for  —  p ;    whence  x  is  found 

r=  U  +  J—  4.  rL      —  =====p^^=E=rr,  the  same  as 
'         ^4  ^271  ,,    .       bd     ,    a'b 

before. 


In  like  manner,  if  things  be  supposed  as  above,  and  there 
be  now  given  z^  +  y'^  =  b ;  then,  by  the  problem  there 
reforred  to^  we  likewise  have  s^  —  5ps^  +  Sp^s  =  b. 

But  the  first  equation,  by  substituting  —  for  its  equal 
y,  becomes  2^+1!=:  b  \    whence  z^^  —  ^2^  ==  —  /% 

z^  =  :^b  +  s/^bb  — "7"%  and  2  =  |/^  +  v  1/;^  —  j^l^  ; 

and    consequently     s     (=     2    +    2/    =    2+-^)     = 

2 

1^  +  \/\bb  —"7^1^  +  —  ^ — ^   =  the  true 

root  of  the   equation  s^  —  5/?5^  +  5p^s  =z  b.       Which 
by  substituting  x  for    6',   and  —  —  for  /?,  gives  x  = 

1^  +  V  kbb  +  STp  —  ==^^^FlT>  fo^  the  true 
root  of  the  equation  x^  +  a>.^  +  ^a^x  ~  ^% 


of  several  Dhnensions.  14r 

Generally^  supposing  z^  +  y^  =  b^  or  z"  +  —  =  ^ 

(because  ^  =  ^),    we    have    z^'^   —    <^^    =    —  /?« ; 

whence  z^  -z^^b  +  s/^bb  - — Ji''\  and  z  = 

;^b  +  V^bb — /^J«  :    therefore  ^  (^  +  t/j  =  2  +  -^  =: 

^3   +  Vic?^;   —  pn\^    +  --^,  ;     which 

^b   +  V  ^bb  +  y'U 
is  the  true  root  of   the    equation  6**  —  npt^^"^  +   n  . 

2  ^  2  3  ^ 

72  —  5      n  —  6      n  —  7        ^o         r^^  \ 

-^—  .  _^—  .  -^—  .  /.^6--«  —  ^c.  (=  2«  +  r) 


This  equation,  by  writing  x  for  6*,  and  —  for  —  /?, 
becomes  .\«  +  aA;^^^  ^  n  ~  S    ^  ^2,  n--4   ^  7^  —  4 


27^  272 

37^  272  372  472 


7  1      •.  .  ^         .  M'         .       ^" 

=    (> ;      and    its    root    a;    =    —    +    v/ —    4-    — 

2  ^4  /^^^ 

-       ■     7  — :-nx*      Wherein  the  two  precedincr 

b      ,       jb^         a"  l«  r  ,    o 


2  ^    4  72' 


theorems  are  included,  with  innumerable  others  of  the 
same  kind ;  but  as  every  one  of  them,  except  the  first, 
requires  a  particular  relation  of  the  coefficients,  seldom 


148  The  Resolution  of  Equatiolis 

occurring  in  the  resolution  of  problems,  I  shall  take  no 
further  notice  of  them  here,  but  proceed  to 

The  Resolution  of  Biquadratic  Equations^  according  to 
Des  Cartes. 

Here  the  second  term  is  to  be  destroyed  as  in  the  so- 
lution of  cubics  ;  which  being  done,  the  given  equation 
\yill  be  reduced  to  this  form,  x^  +  ax^  +  &x  +  c  =  0 ; 
wherein  a^  h^  and  c  may  represent  any  quantities  whatever, 
positive  or  negative.  Assume  ^^  +  /?jv  4-  ^  X  a:^  -f  r;c  +  5 
=  a:^  +  ^^^  -^  hx  -^^  c  ;  or,  which  is  the  same  thing,  let 
the  biquadratic  be  considered  as  produced  by  the  mul- 
tiplication of  the  two  quadratics  x^  +  px  +  y  =  0,  and 
x^  +  rx  +  s  =  O:  then,  these  last  being  actually  mul- 
tiplied into  each  other,  we  shall  have  x"^  +  ax^  +  hx 

+  c  =  .:v'^  +  ^  }►  a;^  +     qK  x^  -{•  P^  \  X  '\'  qs  \    whence, 

^  prj  ^^  ^ 

by  equating  the  homologous  terms  (in  order  to  deter- 
mine the  value  of  the  assumed  coefficients  /^,  ^,  r,  and  ^) 
we  have  p  -j-  r  =  0^  s  +  q  +  pr  =:  a^  ps  +  qr  =z  by 
and  qs  =z  c ;  from  the  first  of  which  r  =  —  p ;  from 
the  second  s  +q  (  =  a  — pr)  =  a  +  p^  ;  and  from  the 

third  s  —  q  z=z  — .      Now,  by  subtracting  the  square  of 

the  last  of  these  from  that  of  the  precedent,  we  have 

4qs  =z  a^  +  2ap^  +  p^  -" ,  that  is,  4c  =  fl^  -f-  2ap- 

-}.  j^4  _^  —    (because    qs  =  c) ;     and  therefore  //   + 

2ap^  _  4.    X  P^  =^^^  i  fr<^n^  which  p  will  be  determined, 

as    in    example  the  second,  of   the  solution  of   cubics. 

Whence  s  {=  ^a  +  ^p^  4 ),  and  5^  (  =  ha  +  ip^  — 

2p 

' — )  are  also  known.      And,  by  extracting  the  roots  ol 
the    two    assumed    quadratics  .^^  -f-  px  -f  ^  ==  O,  and 


of  several  Dimensions.  149 

P 
x^  +  rx  +  s  z=z  O,  we  have  x^  in  the  one,  =  —  —  ± 

^lE  —  q  ;     and,  in  the  other,  =  ■ ±  y  II  —  s 

=  ^  ±  v/ —  —  s^  because  r  =  —  p.      Therefore  the 
four  roots  of  the  biquadratic,  x^  -f  ax^  +  bx  +  c  z=z  Q^ 

T+VT         '2         V4        ^'~  o    +  V4         1^ 


are 

2    " 

p.^JptZ 

^  4, 


and  ■ 

2 


EXAMPLE. 


Let  the  equation  propounded  be  y^  —  42/^  -— .  8y  ^  3^ 
=  0;  then,  to  take  away  the  second  term  thereof,  let 
X  +  1  =  y  'y  whence,  by  substitution,  x^  ^  —  6x^  — - 
16:^  +  21  =  0;  which  being  compared  with  the  ge- 
neral equation,  x^  ^  +  axr^  +  bx  +  c  z=z  0^  we  here 
have  a  =.  —  6,  ^  =  —  16,  and  c  =  21  ;  and  conse- 
quently/^«  —  12/^  —  48/^2  (^_-.  ^6  ^  2ap^  it  4^  I  P"^)  = 

256  (  =  b^).  Now,  to  destroy  the  second  term  of  this 
last  equation  also,  make  2;  +  4  =  />^ ;  and  then,  this 
value  being  substituted,  you  will  have  z^  —  962 
=  576 ;    whence,   by   the    method    above    explained,    z 

will  be  found  (  =  288  +  \J2SsY  —  32]^  ^  + 
32 
=^3       ■  ^-==:=.j)  =  12.       Therefore  p  C- 


288 


+    Vz88T  —  32l^  P 


a  p^  b 


Vz  +  4)  is  =  4,  5  (=  _  +  ^  +  — )  =  3,  and 
5r(=:|.+^^~-)=:7;  consequently  L  + 
^S^s   :.   3,  A  _  JpTZZ  ^U^L  ^ 

^4  2^4  2     ^ 


150  The  Resolution  of  Equations 

^7=-.  +  V—,  and  -p--4inr,^ 

■—2  —  V  —  3  ;  which  are  the  four  roots  of  the  equation 
x"^  —  ^x^  ---,  16^  +  21  ;  to  each  of  which  let  unity  be 
added,  and  you  will  have  4,  2,  —  ^  "%  '^  —  ^5  ^^^  —  ^ 
— -V— 3,  for  the  four  roots  of  the  equation  proposed  j 
whereof  the  two  last  are  impossible. 

And  that  these  roots  are  truly  assigned,  may  be  easily 
proved  by  multiplyinjj^  the  equations,  y  —  4  =  0,  ^  —  2 
=  0,  z/  +  1  —  V —  3  =  0,  and  2/  4.  1  +  V — 3  =  0, 
thus  arising,  continually  together ;  for,  from  thence,  the 
very  equation  given  will  be  produced. 

The  Resolution  of  Biquadratics  by  another  Method* 

In  the  method  of  Des  Cartes^  above  explained,  all  biqua- 
dratic equations  are  supposed  to  be  generated  from  the 
multiplication  of  two  quadratic  ones  :  but,  according  to 
the  way  which  I  am  now  going  to  lay  down,  every  such 
equation  is  conceived  to  arise  by  taking  the  difference  of 
two  complete  squares. 

Here,  the  general  equation  x^  -f-  ax^  4-  bx^  ^  ex  -^  d 
=  0  being  proposed,  we  are  to  assume  x^  -f-  ^ax  +  A"J^ 
—  Ba;  +  CY  =  X*  +  ax^  +  bx^  +  ex  +  d :  in  which 
A,  B,  and  C  represent  unknown  quantities,  to  be  deter- 
mined. 

Then,  x^  -f-  iax  +  A,  and  Bx  +  C  being  actually  involv- 
ed, we  shall  have 

^  ^    +   ^a^x^  +  a  Ax  -f  A^Y  =z  x'^  +  ax^  +  bx^  -f 

^  #   „   B2jc2_2BC;c—  C^J 

ex  +d:  from  whence,  by  equating  the  homologous  terms, 
will  be  given, 

1.  2A  -f.  ia^  _  B2  =  ^,  or  2A  +  ia^  —  b  —     B^  ; 

2.  a  A  —  2B€  =  c,  or  aA  —  c  =  2BC ; 

3.  A^—      C2         =^,orA2  — ^  =C\ 

Let  now  the  first  and  last  of  these  equations  be  multi- 
plied   together^    and    the    product    will,    evidently,    be 


of  several  Dimensions.  151 

equal  to  i  of  the  square  of  the  second,  that  is,  2A^  + 
^aa  —  b  X  A.""  —  2d\  —  d  X  ^aa  —  T  (  =  B^C^)  = 
i.  X  a^^  —  ^«^A  +  c^  (  =  B^C^).  Whence,  denot- 
ing  the  given  quantities  \ac  —  d^  and  \&  +  d^  X  ^aa  —  b 
by  k  and  /,  respectively,  tliere  arises  this  cubic  equation, 
A^  —  ibA.^  +  kA  —  4/  =  0 :  by  means  whereof  the 
value  of  A  may  be  determined  (as  hath  been  already 
taught) ;  from  which,  and  the  preceding  equations,  both 
B  and  C  will  be  known,  B  being  given  from  thence  =r 

V2  A  4-  ^aa  —  6,  an.d  C  =  II~. 

2r> 

The  several  val^es  of  A,  B,  and  C,  being  thus  found, 
that  of  X  will  be  readily  obtained :  for  x^  -f  hax  +  A")^ 
—  Bjv  -f  Cl]^  being  universally,  in  all  circumstances  of 
x^  equal  to  x"^  +  ax^  -f  bx^  -f-  ex  +  d^  it  is  evident, 
that  when  the  value  of  x  is  taken  such,  that  the  latter 
of  these  compressions  becomes  equal  to  nothing,  the  for- 
mer must  likewise  be  =  0  ;  and  consequently 
x^  -f  ^ax  +  A'Y  =  Ba;  4-  Cp:  whence  by  extracting 
the  square  root  on  both  sides,  x^  +  iax  -f-  A  =  -t  B;t^  ±  C  ; 

which,  solved,  gives  x-^-^  .|B — \a  tH  V^^^  ^  ib]"^.  C —  A 
=  ±  |B  —  J«  ±  S/-^^  -f  iah  -f  ib^  di  C  —  "A;  exhibit- 
ing all  the  tour  different  roots  of  the  given  equation,  ac- 
cording to  the  variation  of  the  signs. 

This  method  will  be  found  to  have  some  advantages 
over  that  explained  above.  In  the  first  place,  there  is  no 
necessity  here  of  being  at  the  trouble  of  exterminating  the 
second  term  of  the  equation,  in  order  to  prepare  it  for  a 
solution  :  secondly,  the  equation  A^  —  i^A^  +  kA  —  ^l 
=  0,  here  brought  out,  is  of  a  more  simple  kind  than  that 
derived  by  the  former  method:  and,  thirdly  (which  ad- 
vantage is  the  most  considerable),  the  value  of  A,  in  this 
equation,  will  be  cominenswate  and  rational  (and  therefore 
the  easier  to  be  discovered),  not  only  when  all  the  roots 
of  the  given  equation  are  commensurate^  but  when  they  are 
irrational  2ind  even  impossible;  as  will  appear  from  the 
examples  subjoined. 


1 52  The  Resolution  of  Equations 

Exajn*  1.  Let  there  he  given  the  equation  x^  +  12.x'' — 
17  =  0. 

Which  being  compared  with  the  general  equation 
^4  ^  ^;^3  ^  ijyp.  -|-  cr  +  ^  =  0,  Ave  have  a  =  O, 
Z>  =  0,  c  =  12,  and  ^  =  —  17 :    therefore  >^  (  \ac  —  d\ 

=  17,  /  (ic^  +  dx  iaa  —  ^)  =  36;  and  consequently 
A3  —  ii^'A^  +  kA  —  ^l  z=  A3  +  17A  —  18  =  0;  where 
it  is  evident,  by  bare  inspection,  that  A  =  1.      Hence 

B  (=  V^A  +  iaa  -1)  =  VY,  C  (=  ^^-^)  = 
—  12 


12  _  ^^      n       ^^ 

=  ==  —  3V2  ;  and  ;^  =  ±  -1^2  ±  \/—  if  3^2 


2\/2 ^  2 

=  ±  ^V  2  q:  Y  q=  3 V  2 .       Therefore   the  four 

roots  of  the  equation  are  |V  2  +  W  ~  3V  2  —  ---, 


3\/2  —  — , 
2' 


iV2  _^—  3V2  —  |,  —  iV/2  +  sj 

and  —  -^VT —  y  3V'T —  ---  ;  whereof  the  first  and  se- 
cond are  impossible. 

Exam.  2.  Let  the  equation  given  be  x^  —  6:>^^  —  58^^ 
_.114x  — 11=0. 

Here  a  =  —  6,  ^  =  —  58,  c  =  —  114,  and  d  =  — 11  ^ 
whence  k  (iac  —  d)  =  182,  /  (ice  +  d  X  kaa  —  b)  =. 
2512;  and  therefore  A^  +  29A2  +  182A  —  1256  =  0. 
Where,  trying  the  divisors  1,  2,  4,  157,  ^c,  of  the  last 
term  (according  to  the  method  delivered  on  p.  134)  the 
third  is  found  to  succeed ;  the  value  of  A  being,  therefore, 
=  4.  Whence  there  is  given  B  =  V75  =  5V3, 
90  — 

C    =    —    ^  ^"^  ^y    2Q[\A    X    (=    ±    iB    —    ia    ± 

10\/  3  '  ^ 


of  several  Dimensions.  153 


Exam*  3,  Z^^  M^r^  ^e  ^loti;  proposed  the  literal  equation 
z^  +  2a2'  —  3r«V  —  38a^2  +  a^  =  0. 

This    equation,    by  dividing   the    whole    by  at*,    and 

writing  x  =  — ,    is  reduced  to  the  following   numeral 

one,  x^  +  2x^  —  37^^  —  38^  +1=0.  If,  therefore, 
tf,  b^  c,  and  d  be  now  expounded  by  2,  —  ^7^  —  38,  and 
1,  respectively,  we  shall  here  have  k  (^ac  —  ^)  =  —  20^ 

I  (K^  +  d  X  ^aa  —  ^)  =  399 ;  and,  therefore,  by  sub- 
stituting  these  values, 

A3  +  y  A2  —  20A  —  3|«  =  O. 
or,  2A3  +  37A2  — .  40A  —  399  =  0. 
Which  equation,  by  the  preceding  methods,  will  be  found 
to  have  three  commensurable  roots,  |,  —  3,  and —  19: 
and  any  one  of  these  may  be  used,  the  result,  take  which 
you  will,  coming  out  exactly  the  same.  Thus,  by  tak- 
ing  —  3,  for  A,  we  shall  have  x^  +  x  —  3=  ±V2x 
4a;  +  2  :  but,  if  A  be  taken  =  J,  then  will  x'^  +  x  +  I 
=  ±  VT  X  3:^  +  1 ;  lastly,  if  A  be  taken  =  —  19,  then  xr^ 
+  X  —  19  =  -f-  6\/lO.  All  which  are,  in  effect,  but 
one  and  the  same  equation,  as  will  readily  appear  by 
squaring  both  sides  of  each,  and  properly  transposing; 
whence  the  given  equation  oc^  +  2x^  —  37 :x^  —  38:^  + 
1  =  0,  will,  in  every  case,  emerge.  And  the  same  ob- 
servation extends  to  all  other  cases,  where  there  are  more 
roots  than  one,  it  being  indifferent  which  value  we  use ; 
unless  that  some  are  to  be  preferred,  as  being  the  most 
simple  and  commodious. 

Having  given  the  general  solution  of  biquadratic 
equations  by  the  means  of  cubic  ones,  I  shall  now 
point  out  two  or  three  particular  cases,  where  every 
thmg  may  be  performed  by  the  resolution  of  a  quadra* 
tic  only. 

X 


iS4f  The  Resolution  of  Eqiuttlons 

These  are  discovered  from  the  preceding  equations^ 
2  A  +  i«2  _  ^  _  B^ 

aA— c  =  2BC 
andA^  — ^=    C^: 
wherein,  if  A  be  supposed  =  0,  it  is  plain  that  ia^  — 
^  =  B2,  —  c  =  2BC,  and  — .  ^  =  C2 :    whence  B  =r 

\/^aa  —  ^,  C  =  ^  '        — ■=»  =  V  —  c^,  and  consequent- 
ly rf  =  ^ ;  by  making/=  6  —  \aa. 
4/ 

Therefore,  in  this  case  (wherein  d  =  — r),  the  general 

equation  o^  4-  j^a:  +  A  =  ±  B:^  ±  C,  will  become  :»c*  + 
ifl;c  =  ±  x\/  — -y  -f  V— .^. 

^    But,  z/  B  ^^  supposed  =  0  ;  then  will  2  A  +  i«^  —  6 
»  0,  and  also  c  A  —  c  =  0 ;  whence  A  =  i^  —  |a^ 


3=if=:-l--,  anddiereforeC(  =  VA2  — flr)  =  Vi//— rf: 
so  that  in  this  case  (where  c  =  ^)  the  generd  equation 


becomes  x^  +  hax  +  ^  ==  4-  V^/^/  — '"^ ;  which,  solved, 
gives  x:=z  —  ^a±  ^^TaY  ~  hf±  V\ff—d. 

Lastly,  if  Q  be  supposed  =  O,  then  will  a  A  —  c  =  0, 
sold  A*  —  t/  =  0 ;    consequently  A  =  —  =  V  6^,  and 

B  (  =  V  2 A  +  4^^  —  b)  =  y  —  — /:  therefore,  in  thisi 

case  (where  d  =  — ),  we  shall  have  :^  +  iax  H =  ± 

aa  ci 


»VfZ7. 


From  the    whole  of  which  it  appears,  that  if  c  be 

»  ^,  or  d.  either,  equal  to  ^,  or  to  ~  (/  being  c= 
2  4/  era 


of  several  Dimensions*  155 

b  —  yia) ;  then,  the  roots  of  the  given  equation,  o^  + 
ax^  +  bo(^  +  ex  +  d  z*  O^  may  be  obtained  by  the  reso- 
lution of  a  quadratic,  only. 

Exam.  1.  Let  there  be  given  x^'^  — •  25x^  +  60x  •—  36 

3=0. 

Here  «  =  0,  *  =  •—  25,  c  =  60,  and  a?  =  —  36 ; 

cc 
therefore,  /    (=    —  25)    being    =    _    (=    —    25), 


4^ 


we  have  [by  case  1)  x^  +  ia:v  =  ±  rvV  —  /  =F  V —  d^ 
that  is,  AT^  =  ±  5x  If  6  :   which,  solved,  gives  ^  =  ±  |  A 

VY  r6,  that  is,  :v  =  I  ±  |,  or  A*  =  —  f  =t  ^ :  so  that 
3,  2,  1,  and  —  6  are  the  four  roots  of  the  equation  pro- 
pounded. 

Exam.  2.  Let  there  be  now  given  x^  +  2qx^  +  Zq^x^ 
+  2q^x  —  r*  =  0. 

Then,  a  being  =  2qy  b  =  3^^,    c  =  2^',    and  d  =s 
—  r^,  thence  will  /  (  =  b  —  iaa)  =  2q^^  and  2l  (  =e 

2q^)  =  c ;  and  so  (the  exan^ple  belonging  to  case  2)  we 
have  y  (  =  —  jg  +  \f|^l'  ~  i/  ±  V  i//  —  6^j  =  —  i? 
±  V  —  lyy  ±  V  ^^  +  r\ 

Exam.  3.  Lastly^  suppose  there  tu ere  given  the  equation 

Here,  a  being  =i  —  9,  6=15,  c  =  —  27,  and  c^  =  9, 
it  is  evident  that  —  (  =  9)  =  ^  (  =  9) :    therefore,  bif 

c  /2c 

cas^  3,  we  have  at*  +  iax  +  —  c=  ±  x\j —  +  ^aa  —  b  5 


156  Tha  Resolution  of  Equations 


that  IS,  x^  ^^  Aix  -^  3  (  =  ±  xS/ ^  +  »-ji  —  15)    =  ± 
^x\/  5  I  which,  solved,  chives 
9  ±  -^^  T '»-  'J7>>  ±  vl-vl 

The  Resolution  of  Literal  Equations^  toherein  the 
given  and  the  unknown  quantity  are  alike  af- 
fected. 

Equations  of  this  kind,  in  which  the  given  and  the 
unknown  quantities  can  be  substituted,  alternately,  for 
each  other,  without  producing  a  new  equation,  are  al- 
ways capable  of  being  reduced  to  others  of  lower  dimen- 
sions. In  order  to  such  a  reduction^  let  the  equation^  if  it 
he  of  an  even  dimension^  he  first  divided  by  the  equal  pow- 
ers of  its  two  quantities  in  the  middle  term  ;  then  assume 
a  new  equation^  by  putting  some  quantity  {or  letter^  equal 
to  the  Sinn  of  the  two  quotients  that  arise  by  dividing  tHoSe 
quantities  one  by  the  other  ^  alternately ;  by  means  of -which 
equation^  let  the  said  quantities  be  exterminated ;  whence  a 
numeral  equation  will  emerge^  of  half  the  dimensions  with 
the  given  literal  one* 

But^  if  the  equation  propounded  be  of  an  odd  dimensioiz^ 
let  it  be^  first,,  divided  by  the  sum  of  its  ttvo  quantities^  so 
will  it  become  of  an  even  dimension^  and  its  resolution  will 
therefore  depend  upon  the  preceding  rule* 

Exam.  1.  Let  there  be  given  the  equation  x^-^Aiax^  +. 
Sa^x"^  —  4rt'x  +  a^  =.0. 

»  .  ".  XX         4i%' 

Here,  dividing  by  a^x^^  we  have  — f-  5 

aa  a 

4a    ^   aa         ^   .      xx    ^    aa          ,       '  >^     ,     ^     ,    ^ 
1 =0  (or 1 4x f- h^  = 

.V        XX .  aa        XX  a  x 

0,  by  joining  tiie  corresponding  terms)  ;    and,  by  mak- 

X  a 

ing  2  =  —  +  — ,  and  squaring  both  sides,   we   have 


of  several  Dimensions.  157 

also,    22  =  —    +  2   +   — ,    or  2»  —  2  =    ^  + 

fl!«  :v::c'  aa         :>jx 

Therefore,    by    substituting    these    vakies,    our     equa- 
tion becomes  2^  —  2  —  4z  +  5  =  0,  or  2*  —  42  = 

X  d 

—  3 ;    whence  2  =c  3.      But  —  +  ---  being  =  2,  we 

ax 

have  AT^  — •  'zax  =  —  «^ ;    and  consequently  x  =  \za  ± 

Vl^2^^^~aa  =  4«  X  2  ±  v  22 —  4  =  ^a  X  3  ±  V"5^ 
in  the  present  case. 

Exam.  2.  i^if  ^A^re  he  given  x^  +  4a;v*  ^-  12(2^ a:^  ~ 
12aV  +  4a^x  +  a^  =  0. 

In  this  case,  we  must  first  divide  by  x  +  cr,  and  the 
quotient  will  come  out  x^  -f  2>ax^  —  ISa^x^  -j-  ^a^x  + 
fx*  =  0:     whence,  by  proceeding  as  in  the  former  ex- 

,             ,          XX        aa                  X          a  ^  ^        ^ 

ample,  we  have  —   +.^+3X 1 ^—  15  =  0, 

aa        XX  ax 

or  22-—  2  +  32  —  15  =  O,    and    from    thence  2  = 
x^lr  —  3 


Exam.  3.   Suppose  there  were  given  7x^  —  2&ax^  *— 
26a^x  +  ra^  =  0.  

This,  divided   by  a^x^^    becomes  7  X    —  +  -r  — 

a?  x^ 

0^       c^  X        a 

26  X  -r  +  -5  =  0.    Now,  making,  as  before,  2  = 1 , 

cr       x^  "  a  ,     X 

x^         a^ 
we  have  2^  — •  2  =  —  +  — ;     and,  multiplying  again 

X         a  ,  x^ 

by  2  =  —  +  — ,  we  likewise  have  2^  —  22  =  —   4. 
ax  a^ 

X  a      *     x^  a^      '    ^    *     x^ 


vV  C*  iA.  C* 

+  —    +-7=  -r-    +  z  -\ r;      and,    therefore, 

a  x^  a^  x^ 

X  a- 

~  32    =  ---    +  -T  •     which    values    being    substi^ 

a^  x^ 


tuted  as  above,  our  equation  becomes  7  X  z^  —  32  — 
26   X  2:^  ~  2  =  0,  or  72»  ~  262^  —  2I2  +  52  =  O, 


158  The  Resolution  of  Equations 

Where^  trying  the  divisors  of  the  last  term,  which  are  1, 
2,  4,  13,  £sPc.  the  third  is  found  to  answer;  2,  conse* 
quently,  being  =  4. 

Exam.  4.  Wherein  let  there  be  given  2a:^  — .  IZa^x^  — * 
13a^x2  +  2a*  =0. 

Here,  dividing,  first,  by  at  +  a,  the  quotient  will  be 
2x^  —  2ax^  —   lia^x^    +    llaV  —   lla'^x^  —  2a*  a; 
+  ^Q^    =    Q ;      which,    divided    again    by    a^x^^    gives 
x^     ,    a^         ^      x^    ,    a^  ^^         X  a 

cr         x^  or         x^  ax 

=  0,  that  is.  2  X  z^  —  Iz  —  2  X  2^  —  2  —  llz  +  11 
=  0,  or  2z^  —  22*  —  172  +  15  =  0  (yid.  p.  119)  :  whence 
2  =  3. 

A  literal  equation  may  be  made  to  correspond  with  a 
numeral  one^  by  substituting  a  unit  in  the  room  of  the 
given  quantity  (or  letter)  :  and  equations  that  do  not 
seem,  at  first,  to  belong  to  the  preceding  class  may  some- 
times be  reduced  to  such,  by  a  proper  substitution ;  that 
is,  by  putting  the  quotient  of  the  first  term  divided  by 
the  last,  equal  to  some  new  unknown  quantity  (or  letter) 
raised  to  the  power  expressing  the  dimension  of  the  equa- 
tion.    Thus,  if  the  equation  given  be  2x^  +  24^;^  — •  315;c^ 

2x^ 
+216:>c  + 162  =  0 ;  by  putting =  y\  we  have  x  =.3yi 

whence,  after  substitution,  the  given  equation  becomes 
162jy^  +  6482/^  —  28352/2  +  648t/  -f-  162  =  0:  which 
now  answers  to  the  rule,  and  may  be  reduced  down  to 
2j/4  +  82/^  —  35Z/2  +  8z/  +  2  =  0. 

Of  the  Resolution  of  Equations  by  Approximation 
and  Converging  Series. 

The  methods  hitherto  given,  for  finding  the  roots  of 
equations,  are  either  very  troublesome  and  labo- 
rious, or  else  confined  to  pruticular  cases  ;  but  that  by 
converging  series,  which  we  are  here  going  to  explain, 
is  universal,  extending  to  all  kinds  of  equations;    and^ 


by  Appro^matmi.  15S 

though  not  accurately  true,  gives  the  value  sought,  with 
little  trouble,  to  a  very  great  degree  of  exactness.  When 
an  equation  is  proposed  to  be  solved  by  this  method,  the 
root  thereof  must,  first  of  all,  be  nearly  estimated  (which, 
from  the  nature  of  the  problem,  and  a  few  trials,  may, 
in  most  cases,  be  very  easily  done)  ;  and  some  letter, 
or  unknown  quantity  (as  2)  must  be  assumed,  to  express 
the  difference  between  that  value,  which  we  will  call  r, 
and  the  ti'ue  value  (x)  ;  then,  instead  of  x^  in  the  given 
equation,  you  are  to  substitute  its  equal,  r  ±  2,  and 
there  will  emerge  a  new  equation,  affected  only  with 
z  and  known  quantities ;  wherein  all  the  terms  having 
two  or  more  dimensions  of  2,  may  be  rejected,  as  in- 
considerable in  respect  of  the  rest ;  which  being  done, 
the  value  of  z  will  be  found,  by  the  resolution  of  a  simple 
equation  ;  from  whence  that  oi  x  (  =  r  ±  2)  will  also  be 
known.  But,  if  this  value  should  not  be  thought  suffi- 
ciently near  the  truth,  the  operation  may  be  repeated,  by 
substituting  the  said  value  instead  of  r,  in  the  equation 
exhibiting  the  value  of  z  ;  which  will  give  a  second  cor- 
rection for  the  value  of  x. 

As  an  example  hereof,  let  the  equation  x^  +  10:^^ 
-f-  50x  =  2600,  be  proposed :  thence,  since  it  appears 
that  X  must,  in  this  case,  be  somewhat  greater  than 
10,  let  r  be  put  =  10,  and  r  +  z  =z  x ;  which  value 
being  substituted  for  x^  in  the  given  equation,  we  have 
r^  +  Zr'^z  +  3r2^  +  2'  +  lOr*  +  20r2  +  102^  +  50r 
+  502  =  2600:  this,  by  rejecting  all  the  terms  where- 
in two  or  more  dimensions  of  2  are  concerned,  is  re- 
duced to  r^   +  2,r^z   +  lOr*   +  20r2  +  50r  +  5O2  = 

car^        1  ^        2600  —  7-3  —  lOr*  —  50r 

2600 ;  whence  2  comes  out  = — — 1 

3r2  +  20r  +  50 
=  0.18,  nearly :  which,  added  to  10  (=  r),  gives  10,18 
for  the  value  of  x.  But,  in  order  to  repeat  the  opera- 
tion, let  this  value  be  substitufed  for  r,  in  the  last  equa- 
tion, and  you  will  have  2  =  —  ,000534r  ;  which,  added 
to  10,1"8,  gives  10,1794653,  for  the  value  of  x^  a  second 
time  corrected.  And,  if  this  last  value  be  again  sub- 
stituted for  r,  you  will  have  a  third  correction  of  x  ;  from 
whence  a  fourth  may,  in  like  manner,  be  found;   and 


160  The  Resolutmi  ofEquatiom 

so  on,  until  you  arrive  to  any  degree  of  exactness  you 
please. 

But,  in  order  to  get  the  general  equation  from  whence 
these  successive  corrections  are  derived,  with  as  little 
trouble  as  possible,  you  may  neglect  all  these  terms,  which, 
in  substituting  for  x  and  its  powers,  would  rise  to  two  or  . 
more  dimensions  o\  the  converging  quantity  :  for  as  they, 
by  the  rule,  are  to  be  omitted,  it  is  better  entirely  to  ex- 
clude them  than  to  take  them  in,  and  afterwards  reject 
them. 

Thus,  in  the  equation  jjct^  +  at^  +  ;^  =  90,  let  r  +  z 
be  put  :=  X ;  and  then,  by  omitting  all  the  powers  of 
2  above  the  first,  we  shall  have  r^  +  ^rz  =  x^^  and 
^,3^  3^.2^  ~  x^^  nearly;  which,  substituted  above,  give 
yZ  ^  3^2^  ^  ^2  ^.  2r2  +  ^  +  2  =  90  ;  whence  %  is  found  = 

90 r^ r^  _  r 

.     Therefore,  if  r  be  now  taken  equal 

3r^  4_  2r  +  1  '  ^ 

to  4  (which,  it  is  easy  to  perceive,  is  nearly  the  true  value 

c     \  u  11  u  (       90  —  64—16  —  4         6. 

of  X)  we  shall  have  2  (  = =  — 1  = 

^  ^  48  +  8  +  1  S7 

0.10,  &?c.  which,  added  to  4,  gives  4.1,  for  the  value  of  x^ 

once  corrected  :  and,  if  this  value  of  x  be  now  substituted 

for  r,  we  shall  have  z  {  = )  ==  ,00283  ; 

which,  added  to  4.1,  gives  4.10283,  for  the  value  of  x^ 
a  second  time  corrected. 

In  the  same  manner,  a  general  theorem  may  be  derived, 
for  equations  of  any  number  of  dimensions.  Let  ax^ 
+  <$>^ '»-i  +  ^;v'»-^  +  <3^^"~' +  e^»-^,  ^c.  =  Q,  be  such  an 
equation,  where  72,  a,  ^,  c,  d^  &:c.  represent  any  given 
quantities,  positive  or  negative  ;  then,  putting  r  +  2  =  .r, 
we  have,  by  the  theorem  in  p.  41, 

x^     =2  r^     +  7?r"~^2:,  &c. 
Ar**~^  =  r""^  +  72  —  1  X  r'*"'^2,  &c. 
^n~2  __.  ^n-2  ^jj^  —  2  X  r^'^^z^  &c. 
&c. 

Which  values  being  substituted  in  the  proposed  equations, 


hy  Approximation.  16 J 


it  becomes  ar"^  +  nar^-^z  +  hr''-^  ->r  n  —  1   X  b^'^^H  + 
^^n-2  ^  j^  —  ^  ^  cr'^-s^  +  ^r"*-3  +  ;z  —  3  X  dr^'^'^z^  &c.  =  ' 
Q.     From  which  z  is  found  = 

Q  —  ar""  —  br^"^  —  cr^-^  —  r/r"-^  —  er«-^,  &c. 


nar^''^+  /z- 1  x  br^'^+n-2  x  cr^'^+ ;z-3  X  dr^'^  -f-  /z-  ^  ,<  ^r"-*,  i^c^ 

As  an  instance  of   the  use  of  this  theorem,  let  the 

equation  —  x^  +  300r  =  1000  Be  propounded.      Here 

n  being  =  3,  «  .=  —  1,  3  =  0,  r  =  300,   and   Q  = 

1000,  we  shall,  by  substituting  these  values  above,  have 

1000  4-  r^  —  300r      .         r  •  ,     /       ^  u 

z  =  — :    m  which  (as  it  appears,  by 

inspection,  that  one  of  the  values  of  x  must  be  greater 

than  3,  but  less  than  4)  let  r  be  taken  =  3  ;    and  z  will 

127 
become   =    -^j-    =    0.5,     and    consequently    jjc    ( =    r 

jit  (  o 

+  z)  -=1  Z.S^  nearly.  Therefore,  to  repeat  the  opera- 
tion, let  3.5  be  now  written  instead  of    r,  and  z  will 

7.125 

come  out  =  '- =  —  0,027  ;     which,  added  to 

263.>^5 

3.5,  gives  3.473,  for  the  value  of  x^  twice  corrected. 
And,  by  repeating  the  operation  once  more,  x  will  be 
found  =  3,47296351  ;  which  is  true  to  the  last  fi- 
gure. 

When  the  root  of  a  pure  power  is  to  be  extracted,  or, 
which  is  the  same  thing,  if  the  proposed  equation  be  x^  = 
Q ;  then,  a  being  =  1,  and  b^  c,  J,  &c.  each  =  0  ;    z,  in 

Q  •—  r" 

this  case,  will  be  barely  =  ~ — 3— ;    which  may  serve 

as  a  general  theorem  for  extracting  the  roots  of  pure 
powers.  Thus,  if  it  were  required  to  extract  the  cube 
root  of  10  ;    then,  n  being  =;  3,   and  Q  =  10,  z   will 

10        ■     y^ 

be  = — ;    in  which  let  r  be  taken  =  2,  and  we 

3r2       '  ' 

$hall   have  z  =  —  =  0,16:  therefore  ;c  =  2,16;  from 
12  ' 

Y 


162  The  Resolution  of  Equations 

whence,  by  repeating  the  operation,  the  next  value  of  x 
will  be  found  =  2,1 544. 

The  manner  of  approximating  hitherto  explained,  as 
all  the  powers  of  the  converging  quantity  after  the  first 
are  rejected,  only  doubles  the  number  of  figures  at  every 
operation.  But  I  shall  now  give  the  investigation  of 
other  rules,  or  formulae^  whereby  the  number  of  places 
may  be  tripled,  quadrupled,  and  even  quintupled,  at  every 
operation. 

Let  there  be  assumed  the  general  equation  az  +  h^^ 
tf-  CT?  -f-  t/z^,  &fc.  =  /? ;  2,  as  above,  being  the  converg- 
ing quantity,  and  «,  ^,  c,  d^  &c.  such  known  numbers  as 
arise  by  substituting  in  the  original  equation,  after  the  va- 
lue of  the  required  root  is  nearly  estimated. 

Then,  by  transposition  and  division,    we    shall  have 

p  hz^  CZ^  dz*    cji        r  1  u  •      . 

2  =  -i ,  crc.  irom  whence,  by  reject- 

a         a  a  a 

p 
ing  all  the  terms  after  the  first,  and  writing  q  z=z~^  there 

will  be  give  z  ^=  q  i  which  value,  taking  in  only  one  term 
of  the  given  series,  I  call  an  approximation  of  the  first  de- 
gree, or  order. 

To  obtain  an  approximation  of  the  second  degree,  or 
such  a  one  as  shall  include  two  terms  of  the  series,  let  the 
value  of  2,  found  as  above,  be  now    substituted  in   the 

bz^ 
second  term  — ,  rejecting  all  the  following  ones  ;  so  shall 

2=^ 2.    :=:  q ^,  which  triples  the  number  of 

a         a  ^  a 

figures  at  every  operation. 

For  an    approximation  of   the  third  degree,    let  this 

last  value   of  z  be   now  substituted  in  the   second  and 

third  terms,  neglecting  every  where  all  such  quantities 

as  have  more  than  three  dimensions  of  q :     whence  z 

will  be  had  (=  a ^H ^ ^)  =  a — — ^^ + 

2bb  —  ac    _ 


by  Approximation*  163 

The  manner  of  continuing  these  approximations  is 
sufficiently  evident:  but  there  are  others,  of  the  same 
degrees,  differing  in  form,  which  are  rather  more  com- 
modious, and  whereof  the  investigation  is  also  some- 
what different. 

It  is  evident  from  the  given  equation,  that 

z  =   . 7 — ^    .     ,  3    . —     If,  therefore,  the  first 

a  +  bz  +  cz^  +  dz-^^  &c. 

value  of  z,  found  as  above,  be  substituted  in  the  denomi- 
nator, and  all  the  terms  after  the  second  be  rejected,  we 

shall  have  2  =  — 2-r-  = 2--. ;    which  is  an  approxi- 

a  +  bq       aa  +  bp 

mation  of  the  second  degree. 

But,  if  for  z  you  write  its  second  value,  q ^-^ 

you  will  then  have  x  ( =  ^ ,-    • )  ?r 

a   +  bq  —    — i.  +   cq^ 
a  ^ 

;  being  an  approximation  of  the  third 


bb 
a  +  bq^-'^^'-^c.q- 

degree. 

A      .      t         .  •  b     ^        ''Xbb  '-^  ac       ,  .      , 

A  gam,  by  writmg  q q^  + .  q^  m  the 

^         a    '^  aa 

room  of  2,  and  neglecting  every  where  all  such  terms 

as  have  more  than  three  dimensions  of  q^  you  will  have 

^(= ^ _ ) 

a  +  bq- q^  + q^+cX  q^ q^  +  «? 

:  which  is 


,   J  bb  ^      ^Zb^        2>bc 

a  +  bq  — c  .  q^  ^ Ld  .  q^ 

^        a  ^         aa  a  ^ 

an  approximation  of  the  fourth  degree. 

It  is  observable,  that  the  powers  of   the  converging 

quantity  q^  in  the  former  approximations,  stand,  all  of 

them,  in  the  numerator ;   but  here^  in  the  denominator : 


164  The  Resolution  of  Equations 

but  there  is  an  artifice  for  bringing  them  alike,  into  both^ 
and  thereby  lessening  the  number  of  dimensions,  without 
taking  away  from  the  the  rate  of  convergency* 
To  begin  with  the  approximation  z  = 

— ^  ,  which  is  of  the  third  degree, 

^  a  '■ 

he 

put  ^  = =  the  coefficient  of  the  last  term  of  the 

*  a  b 

denominator  divided  by  that  of  the  last  but  one  \  so  shall 

2:  =  -— /- ;    whereof  the  numerator  and  the 

a  '\'  bq  —  bscf" 

denominator,    being    equally  multiplied    by   1   +  sq^    it 

becomes  z  = r-4 — j — z tt-\  • 

a  A^  bq  —  bsq^   +  asq   +  bsq^  —  bs^q-^ 

but,  the  approximation  being  only  of  the  third  degree, 

hs^cf  may  be  rejected,  and  so  we  have 

p  +  pqs  a  4-  sp  .  p 

Z  =  — ■  -^ '  = — ■■ . 

a  -^-h  -^-as  .  q      aa  -\- b  '\-  as  •  p 
In  the  same  manner,  in  order  to  exterminate  the  third 
dimension  of  q  out  of  the  equation, 

^  = P  

a  +  bq c   .   q^  +  — +  d  .   q^ 

^  a  ^  aa  a 

put  ty  =  —  -I — —  =  the  coefficient  of  the  last  term 

i  a       bb  —  ac 

of  the  denominator  dwided  bv  that  of  the  last  but  one ; 

P 


then  will  z  =: 


a  +hq^^-^  _c  .  /  +  ~ 


a  ^  bq  •—  bsq^  +  bswq 
^  tb 


bb  ^    .    bb  . 

a 

(because  s  -=.  —  —  -7-J  ; 


whereof  the  terms  being  equally  multiplied  by  1  -f-  wq^ 

&c.  we  thence  have  z  ==  — -—7 — —, — 2"— — - — ,  .  ,  o 

a  +  bq  —  bsq^  +  axvq  +  bwtf 


by  Approximation^  .165 


pXH  +toq'  == 

:  which  is  an  approxi- 


i^  -f  ^  -f  aiv  .  q  -^7V  —  s  •  bg 
ap  X  a  -i-  wp 


a  X  aa  +  b  +  axi^  .p-^w  —  s  .  pp 

mation  ot  the  fourth  degree^  and  quintuples  the  number 

of  figures  at  every  operation. 

By  pursuing  the  same  method,  other  equations  might 
be  determined,  to  include  five  or  more  terms  of  the  given 
series  ;  but,  then,  they  w  ould  be  found  more  tedious,  and 
perplexed  in  proportion;  so  that  no  real  advantage,  in 
practice,  could  be  reaped  therefrom.  I  shall,  therefore, 
proceed  now  to  illustrate  what  is  laid  down  above  by  a  few 
examples. 

Exam*  1.     Let  the  equation  given  be  x^  +  20:^  =  100. 

Here,  x  appearing,  by  inspection,  to  be  something 
greater  than  4,  make  4  +  2  =  x  i  then  the  given  equa- 
tion, by  substitution,  becomes  282  +  z^  -^  4,  There- 
fore, in  this  case,    a  =  28,    ^  =   1,    c  =  0,  &c.  and 

A           A'                      .1            ^P         f        112          28. 
f;  =  4  ;    and  consequently  i^ (  =    =  )  = 

^  ^         ^   aa  -^  bp  ^       7%S        19r 

0.14213  ;  which  is  one  approximation  of  the  value  of  z* 

b           c 
But,  if  greater  exactness  be  required,  then  s  ( j-) 

being  here  — ,  and  w  (^ — f-    ,,       — )  =  — ,  we  shall, 
^  28  ^a    ^    bb~ac^         I4'  ' 

according  to  our  two  l2LStformiilce^  have 


Z    (—            a  +  sp  .p              _            28  4-1  X  4  _ 

aa   +  b   +  ds   .  /;           28    X    28   +  2    X  4    ~ 
90    I   X.                   IQ/ 

28   X   7   +    2    =   1386_=    0'14213564,    nearly;  and 

^  r     ap  X  a-^  Tvp _.  28  X  4  X  28  4-  I 

axaa.^£^a^.p+^I^^.pp      28  X  784  +  l¥-f-|- 

~  28  X  28  f  ^  _  28  X  198  _  5544  _ 
7  X  796  +  ^  ""  49  X  796  +  1  "^  39005  "" 


166  The  Resolution  of  Equations 

0.1421356236,  more  nearly ;  which  value  is  true  to  the 
last  figure. 

Exam,  2.  Suppose  the  given  equation^  when  prepared  for 
a  solution^  to  be  7682  +  482^  +  2^  =  —  96. 

In  this  case,  a  =  768,  b  =  48,  c  =;:  1,  c^  =  0,  /?  = 

1  .    '  2b.    ^    ad — be        1  48 

=  — ,  and  T(;  (=  —)  + 


24  ^      a  bb  —  ac        8  48  X  48  —  768 

1  13         ^-        ^  /I  4-  fyns 

=  — .      rherelore 


8  48— 16       32  a-\^b-^as.q 

—  —  96  —  96  X  —  I  X  aV  _  —  95  4-  |  _  —  191  _ 
"~     768  +  48  -I-  3;i  X  —  j  ^^^ — 6—4""      1516    "" 

—  0.1259894,  nearly  ;  or z  = p'\-pnw_ 


a  +  b  +  aw  .q-^-w  —  s  .  bqq 


96  —  96  X  —  i  V  ^\  —  96  -f 


768+48  +  7:^X— i-f/^X  n  768  — 6  — 9  +  ^f^ 
^  -96X128  +  9X16  ^  _  12144  ^  _  o,l259894802, 
753  X  128  +  5  96389  '  • 

more  nearly. 

In  the  same  manner  the  roots  of  other  equations  may 
be  approached  :  but,  to  avoid  trouble  in  preparing  the 
equation  for  a  solution,  you  may  every  where  neglect 
all  such  powers  of  the  converging  quantity,  2,  as  would 
rise  higher  than  the  degree  or  order  of  the  approxima- 
tion you  intend  to  work  b}^  And,  further  to  facilitate 
the  labour  of  such  a  transformation,  the  following  ge- 
neral equations  of  the  values  of  />,  «,  b^  c,  d,  &c.  may  be 
used. 

pz=zk--  ar  —-*     /3r^  —     yr^  —  ^r^,  csPc. 

a=zcc  +  i£(ir  +    Syr^  -f    4^r^^  ^c. 

A  =  ^  +  3rr  +     6h^  +  lOer^  fcPc* 

c  =  y  +  4i'r  -f  lOfr^  +  uPc. 

d=S'  -^oer  -f  &c. 

The  original  equation  being  ax  +  ^x^  +  yx^  +  h.^  +  tx*^ 
E^fc.  =  k  :  from  whence,  by  making  r  +  2  =  ;^,  the  above 
values  are  deduced- 


by  Approximation.  16ir 

The  better  to  illustrate  the  use  of  what  is  here  laid  down^ 
I  shall  subjoin  another  example  ;  wherein  let  there  be 
given  x'^  +  2:c^  +  Sx^  +  4x^  -f  5x  (or  5x  -f  4^:^  +  3;^^  + 
^x*  +  A*)  =  54321  J  to  find  Xj  by  an  approximation  of 
the  second  degree. 

In  this  case,  k  being  =  54321,  «  =  5,  j3  =  4,  y  =  3, 
J^  =  2,  and  e  =  1,  we  have 

p  =z  54321  —    5r  —  4r^  —  3r^  —  2r^  —  r% 

a=z5  +  8r+    9r^  +  8r^  +  5r^,  and 

^  =  4  +  9r  +  12r2  +  lOr^. 
Which    values,    by  assuming  r  =  8,  will  become  p  = 
11529,    a  =   25221,    and    b  =   5964:     whence    <^    ( = 

^1  =  0,45,  and  2  (  =  _A^)  = 11^2. =0,41; 

a^         \  ^      a  +  bq^        25^21+2683 

and  therefore  ;v  (=  r  +  z)  =  8,41,  nearly. 

To  repeat  the  operation,  let  8,41  be  now  substituted  for 
r;    so  shall  p  =  135,92,  a  =  30479,  b  =  6876,  ^  (  = 

t)  =  0.00445,  and  2  (=  — ^)  =  1^?^ = 

a^  ^       a  +  bq'         30479  +  30 

0,004455:   which,  added  to  8,41,  gives  8,414455,  for  the 

next  value  of  x. 

The  formulce^  or  approximations  determined  in  the  pre- 
ceding pages,  are  general,  answering  to  equations  of  all 
degrees  howsoever  affected  ;  but  in  the  extraction  of  the 
roots  oi  pure  powers,  the  process  will  be  more  simple,  and 
the  theorems  themselves  very  much  abbreviated. 

For  let  x^  :=.  k  be  the  equation  whereof  the  root  x  is 
to  be  extracted ;  then,  by  assuming  r  nearly  equal  to  x^ 
and  making  r  x  1  +  z  =  ^,  our  equation  will  become 

r^  X  1  +  z]"  =  ^,  or  1  +  zY  =  — ,'  that  is,  1  +  nz 

n — 1    2   .  n — 1      n  —  2        „    ,  n — 1 

2  2  3  ^  2 

•  — - —   .  .  2*,  £sPc.  =  — :   from  whence,  by  trans- 

3  4  r'* 

position  and  division,  %  +  ^  .  z^  +  ^       ■-  .  ^^       .   .  2} 


168  The  Resolution  of  Equations 

2  3  4  nr"^ 

Here,    by  a  comparison  with    the    general    equation, 

az  +  bz^  +  cz^   +  dz\  ^c.   —  p,    we   have    ^  =   1, 

/.  __  ^  —  1    ^        w — 1      n  —  ^  72  —  1      w  — 2 

0  =  — - — ,  c  = .    ^  = . . 

2  2  3'  2  3 

— -— ^,  £s?c.  and  p  =  ^  :    whence  q  f-^)  =z  p  ;    s 

,^  c.         n  —  1  n —  2        72-4-1  '  2i^ 

/- J  ~  — ^ _-  .    and  w  ( — 

^a  b^  2  3  6  ^  a 

h _72 1        yV.72 2.72 3 I-  .72 1  .72  —  2 

h  —  ^'l^    1  rT^zrt'i:^:      ^ 


_    72   1  72   2    .    72   —    3    2^2   2    .    72   —   2    ___ 

"         1  2  .  n  +  1  "^ 

72  —  1  72  — -2 72    1 

— — -   + -r-  x/2— -3  —  272  +  2=   :; + 

1       ^  2  .  72  +  1  ;  :     ^  1        ■ 

n  —  2  n  —  1  72  —  2        72       ^, 

.=.  X—  n  +  l=:  — j 5—  =  — .     There- 

2  .  ;z  +  1  .  ..  .         . 

fore,  for  an  approximatiSh  of  the  third  degree,  we  have 

a  X  sfi  .  p       ___         t  4-  \t>  .  n  -^  1  .  p         __ 


aa  +  b  +  as  .  p  ^  — 1        22  4-1 

"^        2       "^       6      '  ^ 

p  +  n     '       »  fi —  ,     ^^^  £^^    ^^    approximation  of  the 
1   +  2;/  —  1   .  ^y  

fourth  degree,  z  = ^ = 


a  +  b  4-  aw  .  q  -{-  w  - —  s  •  bq^ 


22  —  1         n  n  //   4  1      n  —  1      ^^ 

^2        ^2^2  6  2  ^ 


by  Approximationi  169 

p  +  hip^  XT 

:  — r    t    2  r ^    HeilCe  it  IS  evi- 

272  1  2n 1      72  —  1 

1+  .p  + •  .p^ 


dent  that  the  root  x  {r  x   1  +  z)  of  the  given  equation 

a-"  =  L  will  be  equal  to  ;*  -(-  ^ ■-    '         hT    nearly  ; 

1  +  2n—l.  ^p 

and  equal  to  r  +  —.Jl2Ll  +  ^^^P 

2??  —  1     ^    .   272  — 1  ,  n  —  1  .  p- 
^        2  ^^  .12 

more  riearhj. 

But  both  these  theorems  will  be  rendered  a  little  more 

727*^ 

commodious,    by    putting    v  =  —- -^  and  substitut- 
ing   — ,    in    the    place  of   its  equal,   /;,    whence,    after 

V 

7'  X  6^     J    f2  -4-  1 

proper  reduction,  x  will  be  had  =  r  -f 


V  X  6v  -f  4?^  —  2 

V          ,          ,           ,                        r  X  2t;  +  72 
nearly;  and  equal  to  r +- — ===== -^ 

'yX2Z;+272 l+^'U 1.272 — 1 

more  nearly* 

I  shall  now  put  down  an  example  or  two,  to  show  the 
use  and  great  exactness  of  these  last  expressions. 

1.,  Let  the  equation  given  be  .v^  =  2,  or,  which  is  the 

same  thing,  let  the  square  root  of  2  be  required. 

Then,  assuming  r  =  1.4,  we  have  ?i  =  2^  A  =  2, 

,     71Y''                2   X   1,96.  „^  ,      ,        p 

V    (^ =    ^ — 1 — )    —    98;     and    therefore    r 

121^^4-     n-fi   ^  ^  MJ<__591    ^   ^       ^ 

^  X   6t;  +  4?2  —  2  9«    X    594 

197  197 

—•    =   1,4  + =   1,41421356;     which    is 

70  X    198  '      ^   13860  ' 

the  value  of  vT  according  to  the  former  approximation  ; 

Z 


irO  The  Resolution  of  Equationa 

but,  according  to  the  latter,  the  answer  will  come  out 

5544 
^••^  +   c^^r..  =  1.41421356236;    which  is  true  to  the 

last  figure  ;  and,  if  with  this  number  the  operation  be  re- 
peated, you  will  have  the  answer  true  to  nearly  60  places 
of  decimals* 

2.     Let  it  be  required  to  extract    the  cube  root  of 
1728.      Here,  taking  r  =  11,  we  shall  have  v  (■— - — ^) 


i  — r« 


3993 
;=  —1-  =  16.05793  ;  and  therefore  r  4- 
397  ^ 


"^'"^-'^  =11,99998, 


2v  +  2n  —  1  X-i)  +  \  X  71' —  1  X  2?2  —  1 

which    differs    from    truth    by   only     part    of    a 

^  ^     50000    ^ 

unit. 

3.    Let  it   be  proposed  to  extract   the   cube    root    of 

500.     Here,  the  required  root  appearing  to  be  less  than 

8,    but  nearer  to  8  than  7,    let  r  be    taken  =   8,   and 

3    V   512 

we  shall  have  v  ( =: )  =  —  128;    and  there- 

^  —12    ^ 

r  .  r  X  "2v  4-71 

fore  r  + 


2v  -i-  2n  —  1   X  V  +  }  X  n  —  1   X  2n  —  1 
=  7.937005259936 ;  which  number  is  true 


96389 
to  the  last  place. 

4.  Lastly,  let  it  be  proposed  to  extract  the  first  sursolid 
root  of  125000.  In  which  case,  k  being  =  125000,  n  =  5^ 
r  =  10,  and  v  =.  20,  the  required  root  will  be  found  =;= 
10,456389. 

Besides  the  different  approximations  hitherto  delivered^ 
there  are  various  other  ways  whereby  the  roots  of  equa- 
tions may  be  approached  ;  but,  of  these,  none  more  gene- 
ral, and  easy  in  practice,  than  the  following  : 


by  Approximation.  171 

Let  the  general  equation,  az  +  bz^  +  cz^  +  dz^  + 
ez*^  cifc.  =  /?,  be  here  resumed;     which,    by  division, 

becomes  z  =  7 -^ .  If, 

P       P  P  P  P 

tjhierefore,  we  make  A  ==  —  j   and  neglect  all  the  terms 

P 

after  the  first,  we  shall  have  z  =  ---  ;  being  an  approxima- 

tion  into  of  the  first  degree. 

And  if  this  value  of  z  be  now  substituted  in  the  se- 
cond term,  and  all  the  following  ones  be  rejected,  we 

shall  then  have  z  =  , —  = =  — - 

a         b         \  CL     K  b  Q 

—  +  — X-T-        _  A  +  — 
p       p       A        p  p 

(by  making  B  = )  ;  which  is  an  approximation  of 

the  second  degree. 

In  order  now  to  get  an  approximation  of  th^  third  de- 
gree, let  this  last  value  be  substituted  in  the  second  term, 
neglecting  all  the  terms  after  the  third  f  so  shall 

z  =  — J T :    but  here,  in  the  room  of 

a         b  A    .     c      ^         '     '       ^ 

p        p  ^        P 

z^^  either  ,of  the  squares  of   the  two  preceding  values 
of  2,  or  their  rectangle  may  be  sub3tituted,  that  is,  either 

-T"  X  -ri  -T7  X  -77>  or  —-  X  T7 ;    but  the  last  of  these 
A       A     B         B  A        B 

(=  — )  is  the  most  commodious  ;    whence  we  have  z  = 

B                    B                 .       ^        aB  +  bA-hc 
r=7r;  supposmg  C  = 

^b  +  Aa+1-     ^  ^ 

p  p  p 

Again,  for  an  approximation  of  the  fourth  degree,  we 

.         b  ^Bc^cBAcA 

have  --2=;~x— ;  ~^=— X-itX-^=  — X— ; 

P  P        ^      P  p        L        B       p        C 

.    d     .         d       B       A       1        d        1  ,.  - 

anci  —  z^  =  ~x  -prX  i^X  --:■  =  — X  —;    which  va- 

fl  p       C        B       A       p       L 


172  The  Resolution  of  Equations 

lues  being  substituted  in  the  general  equation,  and  all  the 
terms  after  the  four  first  rejected,  there  now  comes  out 

rr:  --_  ;  by  makmg  D  = ---. -i— . 

i)  p 

In  like  manner,  for  an  approximation  of  the  fifth  de- 

gree,  we  shall  have  —  ^=:— x— -,  —  2^  =  —  Xtt-X  t- 
p  p       Y>     p  p       B        C 

cB     d    ^         d       C        B        A      dA        ,    e      ^ 

=  /7d>  "  ==  7""  D  "^  CT^  B  =^'  ^"^  7  ' 
C  B  A  1  e  ^ 

D^Z^li^A^pD''  consequently   z 

Z'          P         P           p  P 

h,  =  — -' •     Whence  the  law  oi  con- 

tinuation  is  manifest ;  whereby  it  appears,  that,  if  there  be 

-         ^          a     ^         ciA  •\-  h     ^         aB  -f  bA   4-  c 
taken  A  =  — ,  B  =  3.         C   = -:_ 

P                     P                                 P 
^      <2C  +  ^B  -f-  cA  4-  ^   ^       oD  -^  bC  +  cB  -^  dA  -^  e 
U  z=.  — ,  H.  =  ^^ t 

P  P 

,.     aK+hB+cC+dB+cA+f  ^      a¥+bE+cD+dC+eB+fA+g 

p  p 

,^      ^         ...  1    A    B    C    D    E    F    ^ .     - 

J-fc.  then  will  ^,  _,  ^,  — ,  -g.,  ^,  — ,  £ifc.  be  so  many 

successive  approximations  to  the  value  of  2,  ascending  gra- 
dually from  the  lowest  to  the  superior  orders. 

An  example  will  help  to  explain  the  use  of  what  is  above 
delivered  ;  wherein  we  will  suppose  the  equation  given  to 
be  122  +  62^  +  23  :=2. 

Here  a  =  12,  ^  =  6,  c  =  1,  <^  =  0,  e  =  O,  &fc.  and  /;  =  2  ; 

whence  A  (  =  ^)  =  6,  B  (=  — ^-J  = 

c,o    r   /-  ^  ^^  +  ^^  +  S  -  12  X  39  4-6  X  6  4-1  _ 


I 


by  ApprQximatton.  173 

505    j^   ,      flrC  4-  ^B  +  cA  -f  d^  __  6X  505  4- 6  X  39  +  6 

=  1635,  &Pc. 

A        2 
Therefore,  --  =  —  =  2,  nearly. 
r>        13 

B  78  ; 

C      505  -^ 

C  101  .,; 

=  —  =  2,  Still  nearer. 
D       654        ' 

From  the  same  equations  the  general  values  of  B,  C,  D, 

^c.  may  be  easily  found,  in  known  terms,  independent  of 

each  other. 

1  hus  B  ( = )  =  -r  -f (because  A  =  — j ; 

^       P        P        P\   P    ^  P^ 

,      ^  ,       «B      hK.       c  ^       «^    .   2ab  .    c 
alsoC(=  _  +  _+_)  =  — +  _.  +  ^; 
p         p         p        p^        p^        p 
,  ^  .      aC      bB      cA       d .       fl4       3^2^      ^ac  +  ^(^ 
and  D(=:  —  4 1 J  = 1 -—^ 

P         P        P       P  P^       P^  P" 

•\ ,  &c.     Therefore 

A  _      op 
B  '^'^^p' 


B  _      p  xa^  +  bp 

C  ""  r/^  +  '■J.abp  -\-cp^  '' 


C  __  p  X  a^  -^  ^abp  +  cp^ 

^       a"^  +  Za^bp  +~2ac  -^Jb  . p^  +  dp^ 


D  _  px  a"^  +  Sci^bp  -f-  2ac  +  bb  .p-  +dp^ 

^       a^  +  A^ctbp  +  3ac  +  2>bb  .  ap^  +  be  +  ar/ .  2/>^  +  ^/)^' 
&Pc.  which  are  so  many  different  approximations  to  the  va* 
lue  of  z. 

Thus  far  regard  has  been  had  to  equations  which 
consist  of  the  simple  powers  of  one  unknown  quantity, 
and  are  no  ways  aft'ected,  either  by  surds  or  fractions. 
If  either  of  these  kinds  of  quantities  be  concerned  m 
an  equation,  the  usual  way  is  to  exterminate  them  by 
multiplication,    or    involution    (as    has    been    taught  i» 


174 


The  Resolution  of  Equations 


sect.  IX.)  But  as  this  method  is,  in  many  cases,  very 
laborious,  and  in  others  altogether  impracticable,  espe- 
cially where  several  surds  are  concerned  in  the  same 
equation,  it  may  not  be  amiss  to  show  how  the  method 
of  converging  series  may  be  also  extended  to  these 
cases,  without  any  such  previous  reduction.  In  order 
to  which  it  will  be  necessary  to  premise,  that  if  A  +  B 
represent  a  compound  quantity,  consisting  of  two  terms, 
and  the  latter  (B)  be  but  small  in  comparison  of  the  for- 
mer ;  then  will 

B 


1 
A  +  B 


1 
A 


A^^^A 


2°. 


A  +  B]2  =  A2  + 
1  1  B 


B 


A2i 


^        A*B 


or  A^  + 


^A 


or- 


B 


A+B]!      Ai      2AI      Ak      2AxAi 


4°.  A  +  B]i  =  A3"  +  —i  or  Ai  + 
3A3 


1 

Ai" 


B 


r-. 


A  +  B]i 

A  +  B"|^  =  A^  + 
1  1 


or- 


AJB 
B 


j>,  nearly. 


SAt      At      3A  X  As 

—        Ai    .    Ba' 

3  or  A4  -f 


4A4 
B 


4A 


B 


A  +  Jb]!     Ai      4AI      Ai      4AxAij 


All  which  will  appear  evident  from  the  general  theo- 
rem at  p.  41  :  from  whence  these  particular  equations, 
or  theorems,  may  be  continued  at  pleasure  ;  the  values 
here  exhibited  being  nothing  more  than  the  two  first 
terms  of  the  series  there  given.  But  now,  to  apply 
them  to  the  purpose  above  mentioned,  let  there  be  given 
V  1  4-  a:^  4-  V2  4-  x'^-\-  v'  3  -f  ;f  2  =  10,  as  an  exam- 
ple, where,  x  being  about  3,  let  3  -f  ^  be  therefore  sub- 
stituted for  »v,  rejecting  all  the  powers  of  e  above  the 


by  Approximation*  1  '/5 

first,    as  inconsiderable,    and    then    the    given    equation 

will  stand  thus,  V  10  +  6^  +  Vll   +  6g  +  v"  12  +  "6t* 

.-=  10:_jDut,  by  theorem  2,  V  10  +  67"  will  be  =  VlO 

3V/ 10    X    ^ 
-j ^  nearly;     for,    in  this    case,    A    =    10, 

and    B  =  6^^    and    therefore   A^  +   =   V 10  4- 

2A  ^ 

oVTo  X  e 


lO 


in  like  manner  isVll   +6^  =  Vll  + 


?i:^iiJi-£,  &c.  and  consequently  VlO  +  ^"^^^  ^  ^  4. 
11  _  ^       ^         •^_        ^  10  ^ 

Vn  +  ^^V^^  +  Vli  +  £i:ili5-i  =   10;    which. 
11  12  ' 

contracted,  gives  9.944  +  2.5^18^  =  10;  whence  2.718e 

=   .056,    and    e  =   .0205;     consequently    x    =   3.0205, 

nearly.      Wherefore,  to  repeat  the  operation,  let  3.0205 

+  e  be  now  substituted  for  x  ;  then  will 

V  10.12342    -L    6.Q41g    +    V  11.12342    +   6.041^    + 

V  12.12342   +  6.041t^  =   10;     whence,    ^z/    theorem  2, 
6.041^  , 

V  10.12342   +   — 7= +   V  11.12342   + 

2V  10.12342 

6.041^ 6.041^ 


—    +  V  12.12342   +   — , =   10,    or 

2N/11. 12342  -      ."  -•  2a/ 12.12342  ' 

9,9987814  +  2.7224^  =  10:  from  which  e  comes  out 
=  .000447,  and  therefore  x  =  3.020947  ;  which  is  true 
to  the  last  place. 

Again,  let  it  be  proposed  to  find  the  root  of  the  equa- 

20^  W  ^v   -i-  ^2 

tion  --; +        ^^   ^  "^    =  34.      Put  20  + 

V  lb  -^  5x  +  x^  25 

e  =  X ;    then,  by  proceeding  as  before,  we  shall   havd 

400  4-  20e  20  -f  e   X   V^405    -4-  40<?  ^ ,        , 

-; : ■==-   4 =   34:     but 

V  ji6  +  45e  ^  25 

(J)y  theorem  3)      ■  is   nearly    =  ■  — 

\/^16   +  45t'  V516 


i76  The  Resolutmi  of  Equations 

45e 


,■==-•   and  (bu  theorem  2)   V  4:05   +  AOe  = 
1032    X   1/516 

V405  +    -=:    which  values  bein^^  substituted  above^ 
V405  ' 

our  equation  becomes 


1  4!Se  20  4-€         -^Oe 

400+20^X-7==— -7— ==  +  — I^xV40o  +  ~:=: 

=  34,  that  is,  400  +  20g  x  .044022  —  .001 92e  + 
20  -{-e  X  .804984  +  .0398^  =  34 ;  whence  rejecting  e^* 
&fc.   we  have  1.712e  =  .1915  ;    and  consequently  e  = 

.1118. 

Thirdly,  let  there,  be  given  V  1  —  x  +  V  1  —  2x^  + 
V  1  —  Sx^  =  2.  Then,  if  0.5  -f  e  be  substituted  there- 
in  for  .Y,  it  will  become  VO.5  —  e  +  V  0.5  —  2ef  + 
V 0.625  —  2.25^  =  2;   or  VO.5  —  VO.5  X  e  +  VKs 

_-^  , 2,25^ 

—  V  0.5  X  2^  -f  V  .625  —  — F==  =  2  ;   whence  3.545e 
^  2V.625 

r=  ,204,  ^=.057,  and  at  =  0,557,  with  which  the  opera- 
tion being  repeated,  the  next  value  of  x  will  come  out  = 

.5516.  __^ 

Lastly,  let  there  be  given  1  +  ;c"j^  + 1  -f- x^\~3  -f- 1  +  x^y* 
=  6,5.      Here,  by  writing  3  +  ^  for  x^  and  proceeding 

as  above,  we  shall  have  2  -| f-  lo"]^  -] !^ + 

281^+'     ^  =6.5)  that  is,  6.455  +  1.23^  =  6.5; 

whence  e  =  .036,  and  x  =  3.036. 

It  may  be  observed  that  this  method,  as  all  the  powers 
of  e  above  the  first  are  rejected,  only  doubles  the  num- 
ber of  places,  at  each  operation :  but,  from  what  is 
therein  shown,  it  is  easy  to  see  how  it  may  be  extended, 
so  as  to  triple,  or  even  quadruple,  that  number ;  but 
then  the  trouble,  in  every  operation,  would  tje  increased 
in  proportion,  so  that  little  or  no  advantage  could  be 
reaped  therefrom. 


by  Approxtmatmii  17  f 

Hitherto  we  have  treated  of  equations  which  include 
bne  unknown  quantity,  only.  If  there  be  two  equa- 
^^  tions  given,  and  as  many  quantities  (x  and  y)  to  be  de- 
W  termined,  one  of  those  quantities  must  first  be  extermi- 
nated, and  the  two  equations  reduced  to  one,  according 
to  what  is  shown  in  sect.  9.  But,  if  this  cannot  be 
readily  done  (which  is  sometimes  the  case)  and  the  un- 
known quantities  be  so  entangled  as  to  render  that 
way  impracticable,  the  following  method  may  be  of 
use. 

Let  the  values  of  x  and  y  be  assumed  pretty  near  the 
truth  (v/hich,  from  the  nature  of  the  problem,  may 
always  be  done)  ;  and  let  the  values  so  assumed  be  de- 
noted by  f  and  ^',  and  what  they  want  of  truth  by  s 
and  /,  respectively  ;  that  is,  lety-f-  s  :=  x^  and  g-  -j-  t  =  y  : 
substitute  these  values  in  both  equations,  rejecting  (by 
reason  of  their  smallness)  all  the  terms  wherein  more 
than  one  single  dimension  of  the  quantities  6'  and  t  are 
concerned :  let  all  the  terms  in  the  first  equation,  which 
are  affected  by  .9,  be  collected  under  their  proper  signs, 
and  denoted  by  A^  ;  iu  like  manner,  let  those  affected 
by  t  be  denoted  by  B^ ;  and  those  affected  neither  by 
s  nor  ty  by  Q:  moreover,  let  the  terms  of  the  second 
equation,  wherein  s  and  t  are  concerned,  be  denoted  by 
as  and  bt^  respectively ;  and  let  the  known  terms,  on 
the  right-hand  side  of  this  equation,  or  those  in  which 
neither  s  nor  t  enters,  be  represented  by  y.  Then  the 
equations  (be  they  of  what  kind  they  will)  will  stand 
thus,  A^  +  Bf  =  Q,  and  as  +  bt  =  q.  By  multiplying 
the  former  of  which  by  Z>,  and  the  latter  by  B,  and  then 
vsubtracting  the  one  from  the  other,  we  shall  have  bAs 

—  Bas  =  bOi  —  B7;    and  therefore  5  =    /;  ^ : 

.  Ab  —  ao 

whence  x  (=f+  s)  is  given. 

Again,  by  multiplying  the  former  equation  by  a,  and 

the  latter  by  A,  £s?f.  we  shall  have  aBt  —  Abt  *=  cQ  -- 

Ay,  artd  therefore  t  =  ^Bl"^  =  A^  -  aQ  .    ^^^^^^^ 

Ba  ~  bA       Ab  —  aB 
!/  (=  ,^  -f  0  i^  l^l^ewise  given. 

2  A 


178  The  Resolution  of  Equations 

It  is  easy  to  see  that  this  method  is  also  applicable,  in 
cases  of  three  or  four  equations,  and  as  many  unknown 
quantities  ;  but  as  these  are  cases  that  seldom  occur  in  the 
resolution  of  problems^  and,  when  they  do,  are  reducible 
to  those  already  considered,  it  will  be  needless  to  take 
further  notice  of  them  here  :  I  shall,  therefore,  content 
myself  with  giving  an  example  or  two,  of  the  use  of  what 
is  above  laid  down^ 

le  Let  there  be  given  x^  +  y^  =  10000,  and  x^  —  z/* 
=3  25000  ;  to  find  x  and  y.  Then,  by  writing  f  +  s 
=  x^  g-  -}- 1  =z  2/,  and  proceeding  according  to  the  afore- 
going directions,  we  shall  have  f"^  -f-  4f^s  -f-  g^  +  A^g^t 
=  lOOOO,  and/^  +  Sf'^s  —  ^^  _  5^^  -.  25000,  or 
4f^s  ^-  4g^t  =  10000  — /^  —  g^,  and  Sf'^s  —  5g*t  = 
25000  +  ^^  —  /^ :  therefore,  in  this  case,  A  =  4/^, 
B  =  4^'^  Q  =.  10000  — /^  —  g-^  a  =z  5f\  d  =  — 
5g^^  and  q  =  25000  +  ^^  —  /^^  But  it  appears,  from 
the  first  of  the  two  given  equations,  that  x  must  be 
something  less  than  10,  and  from  the  second  that  y  must 
be  less  than  x  :  I  thei'efore  take  y^=  9,  and  ^  =  8  ,*  arid 
then  A  becomes  =  2916,  B  =  2048,  Q  =  —  65/,  a  = 
32805,  ^  =  —  20480,   ^  =  —  1281  ;    and  therefore  * 

cfJ^p  =  _o.,=,.„,,(^=^,=-o.u. 

hence  x  =  8.87,  and  y  =  7.86,  nearly. 

Therefore,  in  order  to  repeat  the  operation,  let  f  be 
now  taken  =  8.S7,  and  g^  =  7.86  ;  then  will  A  =  2791, 
B  =  1942,  Q  =  _  6.76,   a  z=  30950,  b  =  ^  19083, 

and  ^  =  94  ;    consequently  ^  (  =  -^ ^)  =  .00047, 

and  t  (=    ^f  ~^)  =  —  .00415  ;  whence  x  ==  8.87047, 
Ab  — .  an 

and  y  =  7.85585;  both  which  values  are  true  to  the  last 

figure. 

Example  2.    Let  there  be  given  20;^  +  xy^[^  +  ^x\^ 


xy 


-=12,  and  >/x^  +  y^  +     ,      '   =13.      Here  the 

-^  Vx^  —  y^ 

given  equations,    by  writing  f  +  s  for  x,   and  g  -^  ^ 


by  Approximation.  1 7-9 

ior  t/,  will  become  20f  -f  20^  -f  fg^  -f  2fgt  +  ^p^]"^ 
+    ^/ Sf  +    Hs  =   12,    and    V/^   +  ^  +  2/^^  +  2^'^ 

4-     >  ,>^    '      o         />  =  13  ;  but 

2QA  +  f^^  +  2^"^  +  2/^^  +  ^'^'1^5  t)y  what  is  shown  in 

— . .2 

p.  174,  will  be  transformed  t^WTW]^  +  ^^       '         -= 
3  X  20/  +fg^ 

X  20s  +  2fgt  +  g^s  (supposing  all  the  terms  that  have^ 

more  than  one  dimension  of  s  and  t^  to  be  rejected,  as 

inconsiderable)  ;   also  V/^  ^^^^2/3+  2gt^  is  trans- 


fi+gt         ^  1 


formed  to  V/^  4.  0-2  1  ,..        •  _.  ^^^ — 

^\//2+^2'        s/f2_^^2_^,2fs—2gt 

1                               fs  —  P-t 
to  —=====■  —  ===== ';  —  ;    therefore  our 

equations  will  stand  thus, 

3  X  20/+/^^ 

Vsf  +   -7=  =  12,  and  VP  +  g^  +     /,  ^   + 


^f\  —  t    r  —  g'x  vp  —  ^ 

=  13:  which  equations,  ify  be  assumed  =  5,  and  g 
=  4,  will  be  reduced  to  5.6462  +  .01045  X  36^*  +  40^ 
+  6.3245  +  .6324^  =  12,  and  6.4031  +  781^  +  .625^ 
+  20  +  5^  +  4^  X  .3333  -^  .1852^^  +  .1482?  =  13  5 
whence  1.008^  -f-  .418if  =  .0293,  and  1.59^  —  5.255^ 
=  .0698:  therefore,  in  this  case,  A  =  1.008,  B  = 
0.418,    Q  ==   .0293,    a  =   1.59,   b   =  —  5.255,  and   q 

IjQ B(7 

=  .0698 :   consequently  s  (=  -^ ^)  =  0.305,  and  t 

Ao  —  ao 
-       Ao'  — —  dQ 

(=  -— ^)  =  — *0040j  therefore  x  =  5.0305  and 

At?  —  aB"^  ' 

y  =  3.9960, 


mo  The  Resolution  of 

SECTION  XIIL 

Of  Indeterminate  or  Unlimited  Problems. 

A  PROBLEM  is  said  to  be  indeterminate,  or  unlimited^ 
when  the  equations  expressing  the  conditions  thereof 
are  fewer  in  number  than  the  unknown  quantities  to  be 
determined  -,  such  kinds  of  problems,  strictly  speaking, 
being  capable  of  innumerable  answers :  but  the  answers 
in  whole  numbers,  to  which  the  question  is  commonly  re- 
strained, are,  for  the  general  part,  limited  to  a  determinate 
number  ;  for  the  more  ready  discovering  of  which,  I  shall 
premise  the  following 

LEMMA. 

Supposing  to  be  an  algebraic  fraction,  in  its 

lowest    terms,    x  being  indeterminate,  and  «,  ^,  and  c 
given  whole  numbers ;  then,  I  say  that  the  least  integer^ 

cix  db  b 

for  the  value  of  x^  that  will  ^.Iso  give  the  value  of  — 

C 
an  integer,  will  be  found  by  the  following  method  of  cal° 
culation. 

Divide  the  denominator  (c)  by  the  coefficient  (a)  of  the 
indeterminate  quantity  ;  also  divide  the  divisor  by  the  re- 
mainder^ and  the  last  divisor  again  by  the  last  reinainder  i 
and  so  o?iy  till  a  unit  only  remains. 

Write  down  all  the  quotients  in  a  line ^  as  they  follow^ 
under  the  first  of  "which  write  a  unit^  and  under  the  second 
zvrite  the  first ;  then  multiply  these  tivo  together ^  and  hav* 
ing  added  the  first  term  of  the  loxver  line  {or  a  unit^ 
to  the  product^  place  the  sum  under  the  third  term  of  thq 
upper  line :  multiply^  in  like  mamier^  the  next  two  corres- 
ponding terms  of  the  tzvo  lines  together^  and^  having  added 
the  second  term  of  the  lower  to  the  product^  put  down  the 
result  under  the  fourth  term  of  the  upper  one:  proceed  on^ 
in  this  way^  till  you  have  multiplied  by  every  number  in  the 
upper  line^ 


Indeterminate  Problems.  181 

Then  multiply  the  last  number  thus  found  by  the  absolute 
quantity  (^)  in  the  numerator  of  the  given  fraction^  and 
divide  the  product  by  the  denominator ;  so  shall  the  re- 
mainder be  the  true  value  of  x^  required ;  provided  the 
number  of  terms  in  the  upper  line  be  even^  and  the  sign  ofb 
negative^  or  if  that  number  be  odd  and  the  sign  of  b  af- 
firmative; but^  if  the  number  of  terms  be  even^  and  the  sign 
ofb  crffirmative^  or  vice  versa,  then  the  difference  between 
the  said  remainder  and  the  denominator  of  the  fraction  xoill 
be  the  true  answer. 

In  the  general  method  here  laid  down,  a  is  supposed 
less  than  c,  and  that  these  two  numbers  are  prime  to  each 
other :  for,  were  they  to  admit  of  a  common  measure, 
whereby  b  is  not  divisible,  the  thing  would  be  impossi- 
ble, that  is,  no  integer  could  be  assigned  for  x^  so  as 

aoc  i  b         »         '* 
to  give  the  value  of  an  integer :    the  reason  of 

which,  as  well  as  of  the  lemma  itself,  will  be  explained  a 
little  farther  on :  here  it  will  be  proper  to  put  down  an 
example  or  two,  to  illustrate  the  use  of  what  has  been  al- 
ready delivered. 

Examp.  1.  Let  the  given  quantity  be '^ — • 

Then  the  operation  will  stand  as  follows  : 

87)256(2 

82)87(1  2,  1,  16,     2 

5)82(16  1,  2,     3,  50,   103 

2)5(2  50 

71X  +  10 


1  256)5150(20 

lo  =  Ar, 


Examp.  2/  Given 

ri)89(i 

18)71(3  1,  3,  1 

17)18(1  1,  1,  4,     5 

1  10 

50": 


1 82  The  Resolution  of 

^xanip.  3.  Gwen  • ^ . 

^  450 

3rr)450(l  1,  5,  6 

73)3/7(5  1,  1,  6     37 

12)73(6  250 

1  1850 

74 


Examp*  4.  Given 


450)9250(20 
250 
450 

200  =  AT. 
987;<r  +  651 

1235    .   * 
987)1235(1  1,  3,  1,  48,       1,       1 

248)987(3  1,  1,  4,     5,  244,  249,493 

243)248(1  _^ 

5)243(48  493 

—3)5(1 

2)3(1 

1 


2465 
2958 


1235)320943(259 


7394 

12193" 

1078 

1235 

157  =  X. 

These  four  examples  comprehend  all  the  different  cases 
that  can  happen  with  regard  to  the  restrictions  specified 
in  the  latter  part  of  the  rule.  I  shall  now  show  the  use 
thereof  in  the  resolution  of  problems. 

PROBLEM  I. 

To  find  the  least  whole  number^  which^  divided  by  17, 
shall  leave  a  remainder  of  7 ;  but^  being  divided  by  26,  the 
remainder  shall  be  13. 

Let  X  be  the  quotient,  by  17,  when  7  remains,  or, 
which  is  the  same  thing,  let  17;f  4-  7  express  the  number 


I 


Indeterminate  Problems.  183 

sought;    then,  since  this  member,  when  13  is  subtracted 
from  it,  is  divisible  by  26,  it  is  manifest  that 

J. J—. ,  or  —  must  be  a  whole  number: 

26  '  26 

whence,  by  proceeding  according  to  the  lemina^  x  will  be 
found  =  8  ;  and  consequently  17x  +  7  =.  143,  the  number 
required.     See  the  operation. 
17)26(1 

}7_  1,  1,  1, 

9)17(1  1>  1,  2,  3 

9_  ^ 

8)9(1  18 

r  26 


PROBLEM  II. 

Supposing  9x  -J~  loz/  =  2000,  it  is  required  to  find  all 
the  possible  values  ofx  and  y  in  xvhole  positive  numbers. 
By  transposing  13z/,  and  dividing  the  whole  equation 

u     r.           1                     2000— 13v       ^^^                2  — % 
by  9,  we  have  x  = ^  =  222  —  y  -^ ^  ; 

which,  as  X  is  a  whole  positive  number^ '  by  the  question, 
must  also  be  a  whole  positive  number,  and  so  likewise 

-^ J  from  which  the  least  value  of  y^  in  whole  num- 
bers, will  come  out  =  5  ;  and  consequently  the  corres- 
ponding value  of  X  =  215.  From  whence  the  rest  of 
the  answers,  which  are  16  in  number,  will  be  found, 
by  adding  9  continually  to  the  last  value  of  y,  and 
subtracting  13  from  that  of  x^  as  in  the  annexed  table, 
which  exhibits  all  the  possible  answers  in  whole  num- 
bers. 

x=215;202|189ll76|1631l50ll37|124|lll|98|85|  72|  59|  46|  33!  20/  7 
^=51  14|  23|  32|  4l|  501  59\  68J  77l86195|1041113|122113lll40ll49 
In  the  same  manner,  the  least  value  of  y,  and  the 
greatest  of  x  being  found,  in  any  other  case,  the  rest  of 
the  answers  will  be  obtained,  by  only  adding  the  co- 
efficient of  x^  in  the  given  equation,  to  the  last  value  of  ^/, 
continually,  and  subtracting  the  coefficient  of  y  from  tlie 


184  The  Resolutkn  t)f 

corresponding  Value  of  $c\  Hence  it  follows,  that,  if  tli€ 
greatest  value  of  x  be  divided  by  the  coefficient  of  y,  the 
remainder  will  be  the  least  value  of  x^  and  that  the  quo- 
tient +  1  will  give  the  number  of  all  the  Answers.  But 
it  is  to  be  observed,  that  the  equations  here  spoken  of  are 
such  wherein  the  said  coefficients  are  prime  to  each  other  ; 
if  this  should  not  be  the  case,  let  the  equation  given  be, 
first  of  all,  reduced  to  one  of  this  form,  by  dividing  by  the 
greatest  common  measure. 

PROBLEM  IIL 

To  find  hoxv  many  different  ways  it  is  possible  to  pay 
\00L  in  guineas  and  pistoles  only;  reckoning  guineas  at 
21  shillings  each^  and  pistoles  at  If. 

Let  AT  represent  the  number  of  guineas,  and  y  that 
of  the  pistoles  ;  then  the  number  of  shillings  in  the 
guineas  being  21;c,  and  in  the  pistoles  17y,  we  shall 
therefore  have  21  at  +  17z/  =  2000,  and  consequently  x  = 

^  =  95  4-  — -^ ^ :    which  beinj^  a  whole 

21  ^21  ^ 

number,  by  the  question,  it  is  manifest  that  — 2— 

must  also  be  an  integer  :  now  the  least  value  of  z/,  in  whole 
numbers,  to  answer  this  condition,  will  be  found  =  4,  and 
the  expression  itself  =  3  ;  the  corresponding,  or  greatest 
value  of  X  being  =  92  ;  which  being  divided  by  1 7,  the 
coefficient  of  y  (according  to  the  preceding  note^^  the  quo- 
tient comes  out  5,  and  the  remainder  7 ;  therefore  the  least 
value  of  X  is  7,  and  the  number  of  answers  (  =  5  +  1) 
;=  6  ;  and  these  are  as  follows  : 


X  = 

92 

75 

58 

41     24 

7 

7/=: 

4 

25 
PRO 

46 
>BLI 

67     88 
iM  IV. 

IOC), 

To  determine  whether  it  be  possible  to  pay  100/.  m  gui- 
neas  and  moidores  only;  the  former  being  reckoned  at  2\. 
shillings  each^  and  the  latter  at  27* 

Here,  by  proceeding  as  in  the  last  question,  we  have 


Indeterminate  Problems.  185 

2000  —  27y 


21 


^2\x  +  27y  =  2000 ;  and  consequently  x'  \ 

=  95  —  J/  —*     ■  *^ — :     where,    the    fraction    being   in 

its  least  terms,  and  the  numbers  6  and  21,  at  the  same 
time,  admitting  of  a  common  measure,  a  solution  in 
whole  numbers  {hy  the  note  to  the  preceding  lemma)  is  im- 
possible. The  reason  of  which  depends  on  these  two 
considerations  :  that,  whatsoever  number  is .  divisible  by 
a  given  number,  must  be  divisible  also  by  all  the  divi- 
sors thereof;  and  that  any  quantity  which  exactly  mea- 
sures the  whole  and  one  part  of  another,  must  do  the 
like  by  the  remaining  part.  Thus,  in  the  present  case, 
the  quantity  6?/  ■—  5,  to  have  the  result  a  whole  num- 
ber, ought  to  be  divisible  by  21,  and  therefore  divisible  by 
3,  likewise,  which  is  here  a  common  measure  of  a  and  c  : 
but  6z/,  the  former  part  of  6z/  —  5,  is  divisible  by  3  ;  there^- 
fore  die  latter  part  —  5  ought  also  to  be  divisible  by  3  ; 
which  is  not  the  case,  and  shows  the  thing  proposed  to  be 
impossible. 

PROBLEM  V. 

A  butcher  bought  a  certain  number  of  sheep  and  oxen^for 
which  hepaidlOOL ;  for  the  sheep  hepaidl7  shillings  apiece^ 
and  for  the  oxen^  07iewith  another^  he  paid  7  pounds  apiece; 
it  is  required  to  find  hoiv  many  he  had  of  each  soj't. 

Let  X  be  the  number  of  sheep,    and  y  that  of  the 

oxen;     then,  the  conditions  of  the  question  being  ex- 

pressed  in  algebraic  terms,  we  shall  have  this  equation, 

viz,    17 X    +    140z/  =  2000;     and    consequently    x  = 

2000  — 140v        ^,^         „           4z/— 11  1  .  ,    ,    . 
^  =  117  —  Sy --^ ;     which  bemg 

a  whole  number,  — must  therefore  be  a  whole 

17 
number  likewise :  whence,  by  proceeding  as  above,  we 
find  2/  =  7,  and  x  ■=.  ^0 ;  and  this  is  the  only  answer 
the  question  will  admit  of;  for  the  greatest  value  of  x 
cannot  in  this  case  be  divided  by  the  coefficient  of  2/, 
that  is,  140  cannot  be  had  in  60 ;    and,  therefore,  ac- 

2B 


186  The  Resolution  of^ 

cording  to  the  preceding  note,  the  question  c^n  have  only 
one  answer,  in  whole  numbers. 

PROBLEM  VI. 

A  certain  number  of  men  and  women  being  merry -making 
together^  the  reckoning  came  to  SS  shillings^  toxvards  the 
discharging  ofwhich^  each  man  paid  2>s.  €d.  and  each  wo- 
man \s.  4^d, :  the  question  is^  to  find  how  many  persons  of 
both  sexes  the  company  consisted  of 

Let  X  represent  the  number  of  men,  and  y  that  of 
the  women  ;    so  shall  42x  +  162/  =  396,  or  2\x  -f-  8z/ 

193 ■       21^ 

=  198;    and  consequently  y  = 1-  =  24  —  2x 

8 

5^1%*  — "  6  5^4%* '      6 

' :    whence,  y  being  a  whole  number,  t- 

8  8 

must  likewise  be  a  whole  number ;  nnd  the  value  of 
x^  answering  this  condition,  will  be  found  =  6  ;  and 
consequently  that  of  z/  (=  24  —  12  —  3)  =  9  ;  which 
two  will  appear  to  be  the  only  numbers  that  can  an- 
swer the  conditions  of  the  question ;  because  21,  the  co- 
efficient of  .T,  is  here  greater  than  9,  the  greatest  value 

of  2/. 

PROBLEM  VIL 

One  bought  12  loaves  for  12  pence ^  whereof  some  were 
tivO'penny  ones  ^others  penny  ones^  and  the  rest  farthing 
ones :  xvhat  number  xvere  there  of  each  sort  ? 

Put  X  =  the  number  of  the  first  sort,  y  =  that  of  the 
second,  and  z  =  that  of  the  third  ;  and  then,  by  the  con- 
ditions of  the  question,  we  have  these  two  equations,  viz> 
X  +    y  +  zz=:  12,  and 
8x  +4<y  +  z  =  48. 

Whereof  the  former  being  subtracted  from  the  latter, 

in  order  to  exterminate  z,  we  thence  get  7x  +  2y  =  26y 

36  — —  Yx  X 

and  therefore  y  = —  =  12  —  2x ;    whence 

^3  3 

it  is  evident  that  the  value  of  ;^  =  3,  and  consequently  that 

y  :=  5^  and  z  =  4 ;  which  are  the  numbers  that  were  to  be 

found. 


indeterminate  Problems^  18/ 


PROBLEM  VIII. 

To  find  the  least  integer  possible^  ruhich^  being  divided  by 
28,  shall  leave  a  remainder  of  19  ;  biit^  being  divided  by  19, 
the  remainder  shall  be  15  ;  and^  being  divided  by  15,  the 
remainder  shall  be  !!• 

First,  to  find  the  least  whole  number  that  can  an- 
swer the  two  first  conditions,  let  the  quotient  by  28, 
the  first  of  the  given  divisors,  be  denoted  by  x^  or,  which 
is  the  same  thing,  let  the  said  number  be  expressed  by  28a: 
4-19;  then  this  number,  when  15  is  subtracted  from  it, 

being  divisible  by  1 9,  it  is  manifest  that  ,  or  its 

equal  x  +  -r ,  must  be  an  integer;    from  whence 

the  least  value  of  x  will  be  found  =8  :  and  conse^ 
quently  28^;  +  19  =  243  ;  which  is  the  least  whole 
number  that  can  possibly  satisfy  the  two  first  condi- 
tions. This  being  found,  let  the  least  number  that  is 
exactly  divisible  by  both  the  said  divisors  28  and  19,  be 
now  assumed;  which,  because  28  and  19  are  prime  to 
each  other,  will  be  equal  to  28  X  19,  or  532  :  then, 
since  the  number  required,  by  the  nature  of  the  pro- 
blem, must  be  some  multiple  of  532,  increased  by  243, 
it  is  plain  that  the  said  number  may  be  represented  by 
532;^  +  243;   from  which,  if  11  be  subtracted,  and  the 

532Ar  -I-  232 
remainder  be  divided  by  15,  the  quotient  ( ^ 

z=^  Z5x  +  15  -{ ^ — )  will  be  a  whole    number  by 

7x  4-7 
the  question,  and  consequently  a  whole  niunber 

also ;  from  whence  the  least  value  of  x  will  be  found  = 
14,  and  consequently  that  of  532.r  -f-  243  =  7691  ; 
which  is  the  number  that  was  to  be  found.  In  the  same 
manner,  the  least  number  possible  may  be  found,  which, 
being  successively  divided  by  four  or  more  given  divisors, 
shall  leave  given  remainders. 


i  88  The  Resolution  of 


PROBLEM  IX. 

Supposing  %7x  +  25Q>y  =  15410 ;  to  determine  the  least 

value  ofx^  and  the  greatest  ofy^  in  whole  positive  numbers. 

By  transposition  and  division  we  have 

15410—87^         ^^       87^  —  50         ,  ,       . 

y  z=.  — ^ =60 ^  :    where  the  Irac- 

^  256  256 

tion  being  the  same  with  that  in  examp.  1  to  the  pre- 
mised lemma^  the  required  value  of  x  will  be  given  from 
thence  =  30 ;  from  thence  that  of  y  will  likewise  be 
known.  But  I  shall  in  this  place  show  the  manner  of 
deducing  these  values,  independent  of  all  previous  consi- 
derations, by  a  method  on  which  the  demonstration  of  the 
lemma  itself  depends. 

In  order  to  this,  it  is  evident,  as  the  quantity  ^7x  —  b 
(supposing  b  =  50)  is  divisible  by  256,  that  its  double 
174a;'  —  2b  must  be  likewise  divisible  by  256.  But 
256.V  is  plainly  divisible  by  256 ;  and  if  from  this  the 
quantity  in  the  preceding  line  be  subtracted,  the  re- 
mainder, 82^  -f-  2^,  will  be  likewise  divisible  by  the 
same  number ;  since  xvhatsoever  number  measures  the 
xvhole^  and  one  part  of  another^  must  do  the  like  by  the  re- 
marning  part :  for  which  reason,  if  the  quantity  last 
found  be  subtracted  from  the  first,  the  remainder  5x  —  Zb 
will  also  be  divisible  by  256 :  and,  if  this  new  remain- 
der, multiplied  by  16,  be  subtracted  from  the  preceding 
one  (in  order  to  farther  diminish  the  coefficient  of  :r), 
the  difference  2x  +  50b  must  be  still  divisible  by  the 
same  number.  In  like  manner,  the  double  of  the  last 
line,  or  remainder,  being  subtracted  from  the  preceding 
one,  we  have  x  —  103^,    a    quantity  still  divisible  by 

256  :     but  — ^  =  20  H ;    therefore  x  —  30  must 

256  256 

be  divisible  by  256  ;  and  consequently  x  must  be  either 
equal  to  30,  or  to  30  increased  by  some  multiple  of  256 ; 
but  30,  being  the  least  value,  is  that  required. 

It  may  not  be  amiss  to  add  here  another  example,  to  il- 
lustrate the  way  of  proceeding  by  this  last  method:  where - 

...      987Ar  4-651 
;n  let  us  suppose  the  quantity  given  to  be  — - — ^-^ 


Indeterminate  Problems^r  1 89 

Then,  making  i  =  651,  the  whole  process  will  stand  as 
follows: 

From         -         -  1235a: 

sub.        -  -         -       9^7X'^b 

l.rem.        «         -  24<^x~b 

1.  rem.  X  3  -       744^x~Zb 

2.  rem.        -         -  243:v  +  4/^ 

3.  rem.  -         -  Sx-^sb 

3.  rem.  X  48       -  240.y —  2406^ 

4.  rem.         -         -  3x  -f  244/^ 


5.  rem.     -         -       -  2x — 249^ 


6.  rem.         -         -  x  +  493b ; 

vvhere,  x  being  without  a  coefficient,  let  493 5,  or  its 
equal  320943,  be  now  divided  by  1235,  the  common 
measure  to  all  those  quantities,  and  the  remainder  will 
be  found  1078  ;  therefore  x  +  1078  is  likewise  divi- 
sible by  1235  ;  and  consequently  the  least  value  of  x 
(=  1235  —  1078)  =  157.  The  manner  of  working, 
according  to  this  method,  may  be  a  little  varied ;  it 
being  to  the  same  effect,  whether  the  last  remainder,  or 
a  multiple  of  it,  be  subtracted  from  the  preceding  one, 
or  the  preceding  one  from  some  greater  multiple  of  the 
last.  Thus,  in  the  example  before  us,  the  quantity 
248x  —  ^,  in  the  third  line,  might  have  been  multiplied 
b}^  4,  and  the  preceding  one  subtracted  from  the  product ; 
which  would  have  given  5x  —  5b  (as  in  the  sixth 
line)  by  one  step  less.  If  the  manner  of  proceeding  in 
these  two  examples  be  compared  with  the  process  for  find- 
ing the  same  values,  according  to  the  lemma^  the  gi'ounds 
••f  this  will  appear  obvious. 

PROBLEM  X. 

Supposing  e^f^  andg  to  denote  given  integers  ;  to  deter- 

X  "~"  e  X     -  f 
viine  the  value  of  x^  such  that  the  quaiitities ,  — -^, 

and        -^,  maij  all  of  t herd  be  integers* 


1 90  The  Resolution  of 

By  making  -~ —  =  z/,  we  have  x  =  28z/  +  e ;  which 

vaKie    being    substituted    in    our    second  expression,  it 

becomes  — — ^  ;     which,  as  well  as  y.   is  to  be 

19  ^ 

a  whole    number :     but  — ,^  T,    ~J^  by  making  b  z=z  e 

— f,  will  be  =  2/  H — ^ — ^-—  ;    and  therefore   19//  and 

18//  +  2(^  being  both  divisible  by  19,  their  difference, 
7j  —  2^,  must  also  be  divisible  by  the  same  number  ; 
whence  it  is  evident,  that  one  value  of  y  is  2b  ;  and 
that  2b  +192  (supposing  z  a  whole  number)  will  be  a 
general  value  of  y  ;  and  consequently  that  x  (=  28z/ 
+  e)  =  5322  +  56b  +  (?  is  a  general  value  of  Xy  an- 
swering the  two  first  conditions.       Let  this,  therefore, 

be  substituted  in  the  remaining  expression ^  ;  which, 

by    that    means,    becomes    i ± — ZZK  =  352 

^  '  15 

^3^+. ZUi  (supposing  <3  =  11^  +>  —  ^  =  12e 

—  liy^ — ^).  Here  152  and  142  +  2/3  being  both  di- 
visible by  15,  their  difference  2  —  2/3  must  likewise  be 
divisible  by  the  same  number ;  and  therefore  one  value 
of  2  will  be  2/3,  and  the  general  value  of  2  =  2/3  + 
15w  :  from  whence  the  general  value  of  x  (=  5322  + 
56b  +  e)  is  given  =  79S0io  +  1064/3  +  56^  +  e; 
which,  by  restoring  the  values  of  b  and  /3,  becomes 
,7980w  +  12825^—  11  reo/' —  1064^. 

Now,  to  have  all  the  terms  affirmative,  and  their  co- 
efficients the  least  possible,  let  w  be  taken  ==  —  e  +  2f 
+  g',  whence  there  results  4845^  +  4200/'  +  6916^, 
for  a  new  value  of  x :  from  wl>ich,  by  expounding  e,  f 
and^,  by  their  given  values,  and  dividing  the  whole  by 
7980,  the  least  value  of  x^  which  is  the  remainder  of  the 
division,  will  be  known. 


Indeterminate  Problems^  19> 


PROBLEM  XL 


If5x  +  7z/  +  112  =  224 ;  it  is  required  to  find  all  the 
possible  values  ofx^  y^  and  2,  in  whole  numbers. 

In  this,  and  other  questions  of  the  same  kind,  where 

you  have  three    or   more  indeterminate  quantities,  and 

only  one  equation,    it   will  be   proper,    first  of   all,    to 

find  the  limits  of  those  quantities.       Thus,  in  the  pre- 

,                    .          224 — 7y  —  ll2         ,  , 
sent  case,  because  ^  is  = --^ ,  and  because 

the  least  values  of  y  and  z  cannot,  by  the  question,  be 

less  than  unity,  it  is  plain  that  x  cannot  be  greater  than 

224 7 11  *^*^  . 

. -,  or  41  :  and,  in  the  same  manner,  it  will 

appear  that  y  cannot  be  greater  than  29,  nor  z  greater 

than  19  ;  which,  therefore,  are  the  required  limits  in  this 

T\/r                    •              •           224  — 7z/  — ll2  . 

Moreover,  smce  :v  is  =  — ^ =  45 


case. 


—  y  —  2z  — 


— i — ^— I! —  =  ^  whole  number,  it  is 


5 


manifest  that    ^     — "^ —  must  also  be  a  whole  num- 
5 

ber :     let  2  +  1   be  therefore    considered    as  a  known 

quantity,  and  let  the  same  be  represented  by  b^  and  then 

the  last  expression  will  become     ■       —  ;    from  which, 

by  proceeding  as  above,  we  shall  get  y  =  2b  =  22  +  2  ; 
whence  the  corresponding  value  of  x  comes  out  =  42 
—  52. 

Let  z  be  now  taken  =  1 ;  then  will  a;*  =  37  and  y 
=  4 ;  from  the  former  of  which  values  let  the  coeffi- 
cient of  y  be  continually  subtracted,  and  to  the  latter, 
let  that  of  X  be  continually  added,  and  we  shall  thence 
have  37,  30,  23,  16,  9,  and  2  for  the  successive  values 
of  x-^  and  4,  9,  14,  19,  24,  and  29  for  the  correspond- 
ing values  of  y  ;  which  are  all  the  possible  answers  when 
2  =  L 


192 


The  Resolution  of 


Let  z  be  now  taken  =  2,  then  x  =  32,  and  j/  =  6 ; 
let  the  former  of  these  values  be  increased  or  decreased 
by  the  multiples  of  7,  and  the  latter  by  those  of  5,  as 
far  as  possible,  till  they  become  negative  ;  so  shall  we 
have  39,  32,  25, 18, 11,  and  4,  for  the  successive  values 
of  X  in  this  case,  and  1,  6,  11,  16,  21,  and  26,  for  the 
respective  values  of  t/  :  which  are  all  the  answers  when 
2  =  2.^  . 

Again,  let  z  be  taken  =  3  ;  then,  by  proceeding  as 
above,  the  corresponding  values  of  x  and  y  will  be  found 
equal  to  34,  27,20,  13,6;  and  3,  8,  13,  18,  23,  respec- 
tively ;  and  so  of  the  rest :  whence  we  have  the  follow* 
ing  answers,  being  60  in  numbers. 


z 

-   y 

.oc 

4.    9.14.19.24.29. 

37.30.23. 16. 

9.2. 

2 

1.    6.11.16.21.26. 

39.32.25.18. 

11.4. 

3 

3.    8.13.18.23. 

34.57.20.13. 

6. 

4 

5.10.15.20.25. 

29.22.15.    8. 

1. 

5 

2.    7.12.17.22. 

31.24.17.10. 

3. 

6 

4.    9.14.19. 

26.19.12.    5. 

7 

1.    6.11.16. 

28.21.14.    7. 

8 

3.    8.13.18. 

23.16.    9.    2. 

9 

5.10.15. 

18.11.    4. 

10 

2.    7.12. 

20.13.    6. 

11 

4.    9.14. 

15*    8.    1. 

12 

1.    6.11. 

17.10.    3. 

13 

3.    8. 

12.    5. 

14 

5.10, 

14.    7. 

15 

2.    7. 

9.    2. 

16 

4 

4. 

171. 

6. 

183. 

1  . 

Indeterminate  Problems.  193 

PROBLEM  XII. 

If  \7x  +  19j/  +2I2  =  400;  it  is  proposed  to  jind 
all  the  possible  values  of  x^  ^/,  and  2,  in  whole  positive 
numbers. 

When  the  coefficients  of  the  indeterminate  quanti- 
ties x^  2/,  and  z  are  nearly  equal,  as  in  this  equation,  it 
will  be  convenient  to  substitute  for  the  sum  of  those 
quantities.  Thus,  let  :v  +  ^  +  2  be  put  =  m ;  then, 
by  subtracting  17  times  this  last  equation  from  the  pre- 
ceding one,  we  shall  have  22/  +  42  =  400  —  1 7w  ;  and 
by  subtracting  the  given  equation  from  21  times  the 
assumed  one,  .r  +  2/  +  2  =  7??,  there  will  remain 
4'X  +  2y=i2\m  —  400.  Therefore,  since  y  and  2  can 
have  no  values  less  than  unity,  it  is  plain,  from  the  first 
of  these  two  equations,  that  400  —  1 7m  cannot  be  less 

than  6,  and  therefore  m  not  greater  than ,  or 

23  :  also,  because,  by  the  second  of  the  two  last  equa- 
tions, 21??z  —  400  cannot  be  less  than  6,  it  is  obvious 

that  m  cannot  be  less  than   IL^^  or   19-.    therefore 

21 

19  and  23  are  the  limits  of  m  in  this  case.  These  be- 
ing determined,  let  4x  be  transposed  in  the  last  equation, 
and  the  whole  be  divided    by  2,    and    we    shall    have 

m 
y  z=z  10?n  — '  200  —  2x  +  — ;     which   being  a  whole 

number,  by  the  question,  —  must  likewise  be  a  whole 

number,  and  consequently  7n  an  even  numbeir ;  which, 
as  the  limits  of  in  are  19  and  23,  can  only  be  20,  or  22  : 
let,  therefore,  m  be  first  taken  =  20,  then  y  will  be- 
come =  10  " —  2x  and  2  (m  —  x  —  2/)  =  10  +  x ; 
wherein  x  being  taken  equal  to  1,  2,  3,  and  4,  suc- 
cessively, we  shall  have  y  equal  to  8,  6,  4,  2,  and  2 
equal  to  11,  12,  13,  14,  respectively,  v.hich  are  four 
of  the  answers  required.  Again,  let  m  be  taken  =  22  ; 
then  will  2/  =  31  —  2x^  and  2  =  ;c  —  9  ;  wherein  let 
X  be  interpreted  by  10,  11,  12,  13,  14,  and  15,  suc- 
cess! velv,  whence  z/ will  come  out  11.  9,  7,  5,  3.  and  1 ; 

2C 


19^4 


The  Resotution  of 


and  X  equal  to  1,  2,  3,  4,  5,  and  6,  respectively.  'J'here- 
lore  we  have  the  ten  following  answers  ;  which  are  all  the 
question  admits  of. 


.V  = 

1 

2 

3 

4 

10 

11 

12 

13 

y  — 

8 

6 

4 

2 

11 

9 

7 

5 

11 

12 

13 

14 

1 

2 

3 

4 

14 
3 


15 
1 

6 


PROBLEM  XIII. 


Supposing  7x  +9y  +  23z  =  9999  ;  it  is  required  to 
determine  the  number  of  all  the  answers^  in  positive  inte- 
gers* 

In  cases  like  this,  where  the  answers  are  very  many,  and 
the  liumber  of  them  only  is  required,  the  following  method 
may  be  used. 

In  the  general  equation  ax  -f-  by  -^  cz  ■=  k  (where  a 
and  b  are  supposed  prime  to  each  other)  let  z  be  assumed 
=  0 ;  and  find  the  greatest  value  of  x^  and  the  least  of 
?;',  in  the  equation  ax  -\-  by  z:^  k^  thence  arising ;  de- 
noting them  by  g  and  / :  find,  moreover,  the  least  posi- 
'tive  value  of  ;z,  in  whole  numbers,  from  the  equation 
am  '\'bn-=L  c^  together  with  the  corresponding  value  of 
;;?,  whether  positive  or  negative  ;  then,  supposing  q  tG 
represent  an  integer,  the  general  value  of  x  may  be  ex- 
pressed by  ^  —  bq  —  7722;,  and  that  of  2/  by  I  -}-  aq  ^-^nz  ^ 
as  will  appear  by  substituting  in  the  general  expression 
ax  +  by  -\-  cz^  which  thereby  becomes  ag  —  abq  - — 
arnz  -\-  bl  -{-  abq  —  bnz  -{-  cz  -=1  k^  as  it  ought  to  be, 
because  ag  +  bl  =:  i,  and  all  the  rest  of  the  terms 
destroy  one  another.  And  it  may  be  observed  farther, 
by  the  bye,  and  is  evident  from  hence,,  that  any  two 
corresponding  values  of  7n  and  72,  determined  from  the 
equation  am  -f-  bn  =  c,  will  equally  fulfil  the  conditions 
of  the  general  equation ;  but  the  least  are  to  be  used, 
as  being  the  most  commodious.  As  to  the  limits  of  z 
and  ^,  these  are  easily  determined  ;  the  former  from  the 
original  equation,  and  the  latter  from  the  general  va- 
lue  o£  X ;     by  which  it  appears  that  q  cannot  exceed 

'>  ~    '  ■ ;  w^herein  the  greatest,  or  the  least  value  of  z  is 


to  be  used,  according  as  the  second  term,  after  substitu- 


Indeterminate  Problems.  195 

tion  for  m^  is  positive  or  negative.  But,  besides  this,  there 
is  another  limit,  or  particular  value  of  q  to  be  determined, 
which  is  of  great  use  in  finding  the  number  of  answers. 

It    is    evident,    from    the    given    equations,    that  the 
values  of  x  will  begin    to    be  negative,  when  z  is  so 

increased    as    to    exceed  ^ ?  ;     and   that    those    of 

m 

y  will,    in  like   manner,    become  negative,    when    z    is 

taken    greater    than    —^ — 2  :      therefore,    as    long    as 

©•  —  hq  .  .        I  -\-  ag     .  ,         _ 

^    contmues    c-reater    than (supposmQ-  the 

m  n        ^    ^^         ^ 

value  of  q  to  be  varied)  so  long  will  x  admit  of  a 
greater  assumption  for  z  than  y  will  admit  of,  without 
producing  negative  values ;  and  vice  versa,  By^  mak- 
ing, therefore,  these  two  expressions  equal  to  each  other, 

the  value  of  a  will  be  cfiven  C=  -^ -)  =  — : 

^  ^  ^     a7n+  nh^  c 

expressing  the  circumstance  wherein  both  the  values  of 
X  and  ?/,  by  increasing  2,  become  negative  together. 
But  this  holds  true  only  when  m  is  a  positive  quantity ; 
for,  in  the  other  case,  the  last  term  ( —  mz)  in  the  ge- 
neral value  of  X  being  positive,  the  particular  values  do 
not  become  negative  by  increasing,  but  by  diminishing 
the  value  of  2  ;  it  being  evident,  that  no  such  can  re- 
sult from  any  assumption  for  2,  but  when  y  is  greater 

than  4-. 
b 
To  apply  these  observations    to    the  equation,    7x  -f 
9y  +  232  =  9999,    proposed,    we    shall,    in    the    first 

place,  by  taking  2  =  0,  have  x  =  1428  —  y : 

whence  the  least  value  of  y  is  given  =  5  ;  and  the 
greatest  o.f  x  =  1422.  Again,  from  the  equation  am  -f- 
hn  =  c,  or  77n  ■+•  9n  =  23,    we  have    m  =  3  —  7i  — 

^ — - —  ;   in  which  the  least  positive  value  of  7i  is  given 

=  1  ;  and  the  corresponding  value  of  771  =  2  ;  and  so 
the  general  values  of  x  and  ?/  do  here  become  14St2  -^ 


196 


The  Resolution  of 


9q  —  22,  and  5  +  7^  —  2,  respectively.  From  the 
former    of   which    the    greater    limit  of  q  is  given  = 

1422  —  2  ^^^_  ,p  ng" — ml  .  ,  , 
~— ,  or  157| ;  and  irom  -^ ,  expressmg  the  les- 
ser limit,  we  have  61,  for  the  value  of  ^,  when  the 
least  value  of  x  becomes  equal  to  that  of  y.  These 
limits  being  assigned/  let  q  be  now  interpreted  by  0, 
1,  2,  3,  4,  5,  ^c.  successively,  up  to  61,  inclusive: 
whence  the  number  of  answers,  or  variations  of  y  cor- 
responding to  every  interpretation,  will  be  found  as  in 
the  margin.  From  whence  it  appears  that  the  arith- 
metical progression  4-fll  -j-18-f"25  -f  32,  £sPc.  con- 
tinued to  62  terms,  will  truly  ex- 
press the  number  of  all  the  an- 
swers when  q  is  less  than  62 : 
which  number  is  therefore  given 
=  4  +  61  X  r  -f  4  X  31  =  13485. 
In  all  which  answers  it  is  evi- 
dent, that  x^  as  well  as  z/,  will 
be  positive  (as  it  ought  to  be) : 
because  it  has  been  proved  that 
the  least  value  of  x^  till  q  be- 
comes (  =  '-^ '^)  =  61|,  will  be  greater  than  that 

of  y ;  which  is  positive,  so  far.  But  now,  to  find  the 
answers  when  q  is  upwards  of  61,  w^e  must  have  re- 
course to  the  general  value  of  x ;  which,  in  these  cases, 
by  the  different  interpretations  of  2,  becomes  negative 
before  that  of  y.      Here,  by  beginning  with  the  greatest 

limit,  and  writing  157,  156, 
155,  154,  ^c.  successively, 
in  the  room  of  ^,  it  will  ap- 
pear, that  the  number  of 
answers  wiil  be  truly  ex- 
pressed by  the  series  4  -f  8 
+  13  +  17  +  22,  £sPc.  con- 
tinued to  157  —  61  terms: 
which  terms  being  united  in 
pairs  (because,  in  every  two 
tenns,  the  same  fraction  in  the  limit  of  2  occurs)  the 


7 

y  = 

N.  Ans. 

0 

5  —  2 

4 

1 

12  —  2 

11 

2 

19—2 

18 

3 

26—2 

25 

4 

33  —  2 

32 

is?c. 

^c. 

&fc. 

nq-  —  mL 


9 

X  = 

2JD 

N.  Ans. 

157 

9—22 

4^^ 

4 

156 

18  —  22 

9 

8 

155 

27—22 

13^ 

13 

154 

36  —  22 

18 

17 

153 

45  —  22 

22| 

22 

&c. 

^c. 

^c. 

^c. 

Indeterminate  Problems*  197 

series  12  +  30  +  48  +  &fc.  thence  arising,  will  be  a  true 
arithmetical  progression  j  whereof  the  common  difference 

being  1 8,  and  the  number  of  terms  =  -1- =  48,  the 

sum  will  therefore  be  given  =  20880 :  to  which  adding 
13485,  the  number  of  answers  when  5^  was  less  than  62, 
the  aggregate  34365  will  be  the  whole  number  of  all  the 
answers  required. 

PROBLEM  XIV. 

To  determine  how  many  different  ways  it  is  possible  to  pay 
1000/.  xvithout  using  any  other  coin  than  croxvns^  gui- 
neas^ and  moidores. 

By  the  conditions  of  the  problem  we  have  5.r  +  21z/ 
+  272  =  20000 ;     where    taking    2:  =  0,    ;c    is    found 

:=  4000  —  42/  —  — ,  and  from  thence  the  least  value  of 

2/  =  0  (0  being  necessarily  included  here  by  the  question)  : 
whence  the  greatest  value  of  x  is  given  =  4000.  More- 
over,   from    the    equation  5m  -j-  21n  =•  27,    we    have 

n  —  2 
m  =  5  —  4;^  —  ^~ ;    from  which  ?2  =  2,  and  w  ::^ 

—  3:    so  that  the  general  values  of  x  and  z/,  given  in 

the  preceding  problem,  will  here  become  4000  —  21^ 

+  3z,  and  5q  —  22.      Moreover,  from  the  given  equa- 

.•         .u             .        1'    '       r                      .     u          20000 
tion,  the  greatest  limit  or  z  appears  to  be  = = 

jor  _  mz  __  4000  +  3  X  740 


740 ;    whence  we  also  have 


21 


«^^  1  V     .       r  1    ^  4000 

=  296  =  the;  greatest  limit  ot  q  ;    and  ^  =  = 

^  ^  21 

190,  expressing  the  lesser  limit  of  y,  when  the  value  of 

X,  answering  to  some  interpretations  of  2,  will  become 

negative,    while    those  of   y    still    continue    affirmative. 

To  find  the  number  of  all  these  affirmative  values,  up 

to  the  greatest  limit  of  ^,  let  0,  1,  2,  3,  4,  5,  qI?c.  be 

now  written  in  the  room  of  f/  (as  in  the  margin).    Whence 

it  is  evident  that  the  said  number  is  composed  of  the 


198 


The  Resolution  of 


<1 

^  = 

Quot. 

N.  Ans. 

0 

0  —  22 

0 

1 

1 

5  —  22 

2^ 

3 

2 

10  —  22 

5 

6 

3 

15  —  22 

n 

8 

4 

20—22 

10 

11 

5 

25  —  22 

124 

13 

&?c. 

tfc. 

£57^C. 

^c. 

series   1+3  +  6  +  8  +  11+13,  £sfc.  continued  to 

29/  terms ;  which  terms 
(setting  aside  the  first)  being 
united  in  pairs,  we  shall 
have  the  arithmetical  pro- 
gression 9  +  19  +  29,  &?c. 
where  the  number  of  terms 
to  be  taken  being  148,  and 
common  difference  10,  the 
last  term  will  therefore  be 
1479,  and  the  sum  of  the 
whole  progression  110112: 
to  which  adding  (1)  the  term  omitted,  we  have  110113, 
for  the  number  of  all  the  answers,  including  those  wherein 
the  value  of  x  is  negative  ;  which  last  must  therefore  be 
found  and  deducted. 

In  order  to  this  we  have  already  found,  that  these  ne- 
gative values  do  not  begin  to  have  place  till  q  is  greater 
than  190:  let,  therefore,  191,  192,  193,  t^c.  be  sub- 
stituted, successively,  for  q ; 
from  whence  it  will  appear 
that  the  number  of  all  the 
said  negative  values  is  truly 
exhibited  by  the  arithmeti- 
cal progression  4  +  11  + 
18  +  25,  ^c.  continued 
to  296  — -190  terms  ;  where- 
of the  sum  is  393/9 ;  which 
subtracted  from  110113,  found  above,  leaves  /0/34,  for 
the  number  of  answers  required. 

After  the  manner  of  these  two  examples  (which  il- 
lustrate the  two  different  cases  of  the  general  solution, 
given  in  the  preceding  problem)  the  number  of  answers 
may  be  found  in  other  equations,'  wherein  there  are 
three  indeterminate  quantities.  But,  in  summing  up 
the  numbers  arising  from  the  different  interpretations 
of  ^,  due  regard  must  be  had  to  the  fractions  exhibited 
in  the  third  column  expressing  the  limits  of  2  ;  because, 
to  have  a  regular  progression,  the  terms  of  the  series  in 
t\\Q  fourth  column,    exhibiting  the  number  of  answers^ 


7 

X. 

Quot. 

N.  Ans. 

191 

32—11 

H 

4 

192 

32  —  32 

10| 

11 

193 

32  —  53 

i/| 

18 

194 

32-/4 

24| 

25 

^c. 

£5?c. 

£5fc. 

isPc. 

Indetermbiate  Problems.  199 

itvust  be  united  by  twos,  threes,  or  fours,  &Pc.  according  as 
one  and  the  same  fraction  occurs  every  second,  third,  or 
fourth,  &fc.  term  (the  odd  terms,  when  there  happen  any 
thing  over,  being  always  to  be  set  aside,  at  the  beginning 
of  the  series).  And  it  may  be  observed  farther,  that,  to 
determine  the  sum  of  the  progression  thus  arising,  it  will 
be  sufficient  to  find  the  first  term  only,  by  an  actual  addi- 
tion :  since  not  only  the  number  of  terms,  but  the  com- 
mon difference  also,  will  be  known  ;  being  always  equal 
to  the  common  difference  of  the  limits  of  z  (or  of  the  quo- 
tients in  the  said  third  column)  multiplied  by  the  square  of 
the  number  of  terms  united  into  one  ;  whereof  the  reason 
k  evident.  But  all  this  relates  to  the  cases  wherein  the 
coefficients  of  the  indeterminate  quantities,  in  the  given 
equations,  are  (tw  o  of  them  at  least)  prime  to  each  other : 
I  shall  add  one  example  mqre,  to  show  the  way  of  proceed- 
ing when  those  coefficients  admit  of  a  common  measure. 


PROBLEM  XV. 

Supposing  \%x  -^^  I5y  +  20z  =  100001  ;   it  is  required 
to  find  the  number  of  all  the  answers  in  positive  integers. 

It  is  evident,  by  transposing  202  and  dividing  by  (3) 

the  greatest  common  measure  of  x  and  z/,  that  Ax  +  5z/, 

22; 2 

and    consequently    its    equal    33333  —  62  ■ —  , 

must  be  an  integer,  and  therefore  22  —  2  divisible  by 
3  :  but  32  is  divisible  by  3,  and  so  the  difference  of 
these  two,  which  is  2  -f  2,  must  be  likewise  divisible 
by  the  same  number,  and  consequently  2  =  1+  some 
multiple  of  3.  Make,  therefore,  1  +  32^  =  2  (ti  be- 
ing an  integer)  ;  then  the  given  equation,  by  substitut- 
ing this  value,  will  become  12x  +  \5y  +  60ic  +  20 
=  100001  ;  which,  by  division,  fcPcv  is  reduced  to 
4x  +  5y  +  20u  =  33327  :  wherein  the  coefficients  of  .v 
and  y  are  now  prime  to  each  other,  and  we  are  to  find 
the  number  of  iill  the  variations,  answering  to  the  dif- 
ferent interpretations  of  ?/,  from  0  to  the  greatest  lim^l 
inclusive. 


200 


The  Resolution^  fcf'c. 


By  proceeding,  therefore,  as  in  the  aforegoing  cases^ 


we  have  x  =  8331  —  y  —  -■ 


whence  the  least  va- 


lue of  z/  is  given  =  3,  and  the  greatest  oi  x  =z  8328. 
Moreover,  from  the  equation  4;?z  +  Sii  =  20,  we  have 


mz=z5' 


•  —  ;  whence  n  =  0,  and  m  z=  3. 
4 


Therefore 


the  general  values  of  x  and  y  (given  in  problem  1 3)  do 

here  become  8328  —  5q  —  5w,  and  3  +  4^' ;  from  the  for- 

8328 
mer  of  which  the  greatest  limit  of  q  is  given  = = 

1 665.  Now,  since  the  value  of  y  will  here  continue  posi- 
tive, in  all  substitutions  for  q  and  w,  as  no  negative  quanr 
tity  enters  therein,  the  whole  numbers  of  answers  will  be 
determined  by  the  values  of  x  alone. 

In  order  to  this,  let  q  be  successively  expounded  by 

1665,  1664,  1663,  ^c. 
and  it  will  thence  appear 
that  the  said  number  will 
be  truly  defined  by  1666 
terms  of  the  arithmetical 
progression  1  -f  2  -f  3 
+  4  +  5,  £s?c.  whereof 
the  sum  is  found  to  be 
1388611. 

When  there  are  four  indeterminate  quantities  in  the 
given  equation,  the  number  of  all  the  answers  may  be  de- 
termined by  the  same  methods  ;  for  any  one  of  those  quan- 
tities may  be  interpreted  by  all  the  integers,  successively," 
up  to  its  greatest  limit,  which  is  easily  determined ;  and 
the  number  of  answers  corresponding  to  each  of  these  in- 
terpretations may  be  found  as  above  ;  the  aggregate  of  all 
which  will  consequently  be  the  whole  number  of  answers 
required :  which  sum  or  aggregate  may,  in  many  cases, 
be  derived  by  the  methods  given  in  section  14,  for  sum- 
ming of  series  by  means  of  a  known  relation  of  their  terms. 
But  this  being  a  matter  of  more  speculation  than  real 
use,  I  shall  now  pass  on  to  other  subjects. 


? 

X 

Quot. 

N.  Ans. 

1665 
1664 
1663 

3 —  5ii 

8  —  5u 

13  — 5u 

H 

1 
2 
3 

The  Jmestigation^  ifc.  ^Oi 


SECTION  XIV, 

The  Investigation  of  the  Sums  of  the  Powers  ofNwn- 
bers  in  Arithmetical  Progression^ 

BESIDES  the  two  sorts  of  progressions  treated  of  in 
section  10,  there  are  infinite  varieties  of  other  kinds  ;  but 
the  most  useful,  and  the  best  known,  are  those  consisting 
of  the  powers  of  numbers  in  arithmetical  progression ; 
such  as  12  +  2^  +  32  +  4^  •  .  .  .  w%  and  1^  +  2^  +  3*  +  4^ 
•  •  •  •  n^,  &fc,  where  n  denotes  the  number  of  terms  to 
which  each  progression  is  to  be  continued*  In  order  to 
investigate  the  sum  of  any  such  progression,  which  is  the 
design  of  this  section,  it  will  be  requisite^  first  of  all,  to 
premise  the  following 

LEMMA. 

If  any  expression,  or  series,  as 

ers  of  an  indeterminate  quantity  72,  be  universally  equal  to 
nothing,  whatsoever  be  the  value  of  n  ;  then,  I  say,  the 
sum  of  the  coefficients  A  —  a,  B  —  ^,  C  — •  c,  &c.  of  each 
rank  of  homologous  terms,  or  of  the  same  powers  of  ;2, 
will  also  be  equal  to  nothing. 

For,  in  the  first  place,  let  the  whole  equation 
An  +  B/z^  +  Cii"  &c.  1         r.   u      V  'A    ^  u  1 

_  an  ^bn^  -  en'  &c.  j  =  ^'  ^^  ^^^^d^d  by  n,  and 

this  being  universally  so,*"  be  the  value  of  n  what 
it  will:     let,    therefore,  n  be  taken  =  0,    and  it    will 

become    4   1=0;     which    being    rejected    as 

such,  out  of  the  last  equation,  we  shall  next  have 
+  Bn  +  Cn^  +  D723  &C.1  ^  ,  ,.  .,. 

^bn   ^cn^~   dn'  Scc.J     =    ^'     whence,    dmdmg 

2D 


iO^  The  Investtgation  of 

again  by  n,  aiid  proceeding  in  the  very  same  manner,  B .—  ^ 
is  also  proved  to  be  =  0  ;  and  from  thence  C  »— .  c,  D  —  d^ 
he.     Q.  E.  D. 

Now,  to  apply  what  is  here  demonstrated  to  the  pur- 
pose above  specified,  it  will  be  proper  to  observe,  first, 
that    as    the   value  of  any  progression,   1^   +  2^   +  3^ 

+  4* 72^,  varies  according  as  72,  the  number  of 

its  terms,  varies,  it  must  (if  it  can  be  expressed  in  a  ge- 
neral manner)  be  explicable  by  7z,  and^  its  powers  with 
determinate  coefficients ;  secondly,  it  is  obvious  that 
those  powers,  in  the  cases  above  proposed,  must  be  ra- 
tional, or  such  whose  indices  are  whole  positive  num- 
bers ;  because  the  progression,  being  an  aggregate  of 
whole  numbers,  cannot  admit  of  surd  quantities  ;  lastly, 
it  will  appear  that  the  greatest  of  the  said  indices  can- 
not exceed  the  common  index  of  the  progression  by  more 
than  unity ;  for,  otherwise,  when  n  is  taken  indefinitely 
great,  the  highest  power  of  ii  would  be  indefinitely  greater 
than  all  the  rest  of  the  terms  put  together. 

Thus,  the  highest  power  of  n,  in  an  expression  univer- 
sally exhibiting  the  value  of  1^  +  2^  -f-  3^  .  .  .  .  •  w^, 
cannot  be  greater  than  n^  ;  for  1^  -f  2^  +  3^  .  .  .  .  .  n^ 
is  manifestly  less  than  n^  (or  n^  +  n^  +  n^  +  &c.  con- 
tinued to  n  terms)  ;  but  7i^,  when  n  is  indefinitely  great, 
is  indefinitely  greater  than  n^^  or  any  other  inferior 
power  of  72,  and  therefore  cannot  enter  into  the  equation. 
This  being  prertiised,  the  method  of  investigation  may 
be  as  follows. 

Case  1°.  To  jind  the  sum  of  the  progression  1  -f  2;  -f-  3 
-f-  4  ....  72. 

Let  Atz^  -|-  B72  be  assumed,  according  to  the  foregoing- 
observations,  as  a  universal  expression  for  the  value 
ofl+2  +  3-f4.  .  .  .  .  72;  where  A  and  B  represent 
unknown,  but  determinate  quantities.  Therefore,  since 
the  equation  is  supposed  to  hold  universally  true,  whatso- 
ever be  the  number  of  terms,  it  is  evident,  that,  if  the  num- 
ber of  terms  be  increased  by  unity,  or,  which  is  the  same 


the  Sians  of  Progressio7is,  3K)3 

tiling,  if  72  +  1  be  written  therein,  instead  of  if^  the  equaT 
iity  will  still  subsist,  and  we  shall  have  A  X  ^  +  l]^  + 

B  X  «  +  1  =  1+2  +  3+4 n  X  n  +  1. 

From  which  the*first  equation  being  subtracted,  there  re- 
mains A  xn  +  if  —  An^  +  B  xn  +  \  —  Bn  =  /z  +  1 : 
this  contracted  will  be  2 An  +  A  +  B  =  72  +  1:  whence 
we  have  2 A  —  Ix^  +  A  +  B  —  1=0:  wherefore, 
by  taking  2 A  —  1=0,  and  A  +  B  —  1  =  0  {accord- 
ing to  the  lemma) ^  we  have  A  =  I,  and  B  =  | ;  and 
consequently  1+2  +  3+4 72(=  An^  +  B^z)  = 

72^         n  72  X  w  -f-  1^ 

- — I ,  or . 

2^2'  2 

Case  2*'«  To  find  the  sum  of  the  progression  1^  +  2^  + 
3^ 72%  or  1  +  4  +  9  +  16 n^. 

Let  A72^  +  B/2^  +  Cw,  according  to  the  aforesaid  ob* 
servations,  be  assumed  =  1^  +  2^  +  3^  +  4^  .  .  .  .  72^ : 
then,  by  reasoning  as  in  the  preceding  case,  we  shall  have 
A  xlT+lT  +  B  xT+TT  +  C  Xn  +\  =  12  +  22  + 
32+4^  .  .  .  .  72^  +  72  +  1]^ ;  that  is,  by  involving  72  +  1 
to  its  several  powers,  A72^  +  3A722  +  3A72  +  A  +  ^r^ 

+  2B)2  +  B  +  C72   +   C    =    1^    +   22    +    32    +  42   .    •    .    .  72^ 

+  72  +  l"]^:    from  which,  subtracting  the  former  equa- 


^  In  this  investigation,  it  is  taken  for  granted,  that  the 
sum  of  the  progression  is  capable  of  being  exhibited  by 
means  of  the  powers  of  72,  with  proper  coefficients : 
which  assumption  is  verified  by  the  process  itself :  for  it 
is  evident  from  thence,  that  the  quantities  A72^  +  B;2, 
and  1  +  2  +  3  +  4  ...  72,  under  the  values  of  A  and 
B  there  determined,  are  always  increased  equally,  by 
taking  the  value  of  72  greater  by  a  unit:  if,  therefore, 
they  be  equal  to  each  other,  when  72  is  =  0  (as  they 
actually  are),  they  must  also  be  equal  when  72  is  1  ;  and 
so,  likewise,  when  72  is  2,  £s?c.  £s?c.  And  the  same  reasoning 
holds  good  in  all  the  following  cases. 


U04f  The  Investigation,  of 

tion,  we  get  ZKii^  +  3Aw  +  A  +  2^n  +  B  +  C 
( =  >  -f  Ip)  =  n^.  +  2)1  +  1;  and  consequently 
?A  — 1  X  ^^^  +  3A  +  2B  —  2  X  ?2  +  A  +  B  +  C  —  1 
=  0;  whence  (^bif  the  lemmd)^  3A  — r  1  =  0,  3A 
+  2B  ~  2  =  0,  and  A  +  B  +  C  —  1  =  0 ;    therefore 

A  =:i-    B  =  -  —  ^^  ^i_    c  =  l  — A  — B=i-- 

3'  3  ^'  6  ' 

and  consequently  1+4  +  9  +  16 n2=:~  + 

,    n         72  .  n  4-  1  .  2?2  +  1 
6'  6 


3    '    2 


Case  3°.  To  determine  the  mm  of  the  progression  P  +2- 
+  33  +  43  ^ 71%  or  1  +  8  +  27  +  64. .  .  .  .  ,  n'. 

By  putting  A/z'*  +  Bn^  +  Ctz^  +  D;z  =  1  +  8  +  27 
+  64  .•,...  72^,  and  proceeding  as  above,  we  shall  have 
4A723  +  eAn^  +  4A72_+  A  +  3B?22  +  3B/z  +  B  +  2C;? 
+  C  +  D  (=  n_+  lY)  =  n^  +  3y;^  +  37Z  +  1  ;  and 
therefore  4A  — "1  x  n^  +  6A  +  3B  —  3  X  ^^^  + 
4A  +  3B  +  2C1II1  X  7i  +  A  +  B+  C  +D  —  1=0: 

1     r>  /      3—- 6A.        1    ^,     3— 4A  — 3B. 
hence  A  =  -,  B  (=  _^)  =  _,C  (= ^ )  • 

=  — ,  D   (=1  —  A  —  B  —  C)  =  0;    and  therefore 


n^         71^ 


or 


1^    +    23    +    3^    +    43 72^=: f-    _    +    — , 

^^  424 

. Z — L.     In  the  verv  same  manner  it  will  be  found 

4 

that 

^A  r^A  ,  A  ^^^  W'*  ^^^  ^^ 

^  5        2    '    3        30 

r       ^r  ^       n^       n^       5n^      n^ 

^       ^  6         2   ^  12        12^ 

72^          72^          72*          72^           72 
:^6  +  2^  +  3^  .  4  .  •  72^  =  —  +  —  + + 

^^^  7^22  6^  42 


the  Sums  of  Progressions.  205 

In  order  to  exemplify  what  has  been  thus  far  deliver- 
ed, let  it,  in  the  first  place,  be  required  to  find  the  sum 
of  the  series  of  squares  1+4  +  9  +  16,  &fc.  continued 
to  10  terms  :    then,  by  substituting  10  for  h,  in  the  ge- 

n  .  w  +  1  .  271  +  1    ,      n^    ,  72^        71  ^ 
neral  expression (ox  ~  +  _  +      ), 

found  by  case  2°,  there  will  come  out  385,  for  the  re- 
quired sum  of  the  progression :  which,  the  number  of 
terms  being  here  small,  may  be  easily  confirmed,  by  ac- 
tually adding  the  10  terms  together.  Secondly,  let  it 
be  required  to  find  the  number  of  cannon-shot  in  a 
square  pile,  whose  side  is  50.     Here,  by  writing  50  for  7i 

7Z  .  72  +  1  .  2/1  +  1  •    1    11  t 

m  the  same  expression,  — ,  we  shall  have 

^50  X   51   X   101>^    42925,    expressing    the    number   of 

shot  in  such  a  pile.  Lastly,  suppose  a  pyramid  composed 
of  100  stones  of  a  cubical  fi.gure  ;  w^hereof  the  length  of 
the  side  of  the  highest  is  one  inch  ;  of  the  second  two 
inches  ;  of  the  third  three  inches,  i^c.  Here,  by  writing 
100  instead  of  tz,  in  the  third  general  expression,  we  have 
25502500,  for  the  number  of  solid  inches  in  such  a  pyra- 
mid. 

Hitherto,  regard  has  been  had  to  such  progressions  as 
have  unity  for  their  first  term,  and  likewise  for  the 
common  difference  ;  but  the  same  equations,  or  theo- 
rems, with  very  little  trouble,  may  be  also  extended 
to  those  cases  where  the  first  term,  Knd  the  common 
difference,  are  any  given  numbers,  provided  the  for- 
mer of  them  be  any  multiple  of  the  latter.  Thus,  sup- 
pose it  were  required  to  find  the  sum  of  the  progression 
6^  +  8^  +  10*  £ffc.  (or  ^,Q>  +  64  +  100  Ssfc.)  conti- 
nued to  eight  terms :  then,  by  making  (4)  the  square  of 
the  common  difference  a  general  multiplicator,  the  given 
expression  will  be  reduced  to  4  X  3^  +  4^  +  5^  ....  10'  : 
but  the  sum  of  the  progression  P  +  2*  +  3*  +  4^  .  .  .  .  10^ 
is  found,  by  the  second  theorem,  to  be  385 ;  from 
which,  if  (5)  the  sum  of  the  two  first  terms  (which  the 


206  The  Iwoestigatiqn  of 

series  3^+4^  +  52 lo^  wants)  be  taken  away,  the 

remainder  will  be  380 ;  and  this,  multiplied  by  4,  gives 
1 520,  for  the  true  sum  of  the  proposed  progressirn  :  and 
so  of  others. 

But  if  the  first  term  be  not  divisible  by  the  common  dif- 
ference, as  in  the  progression,  5^  +  7^  +  9^  &Pc.  the  specu- 
lation is  a  little  more  difficult ;  nevertheless,  the  sum  of 
the  series,  in  any  such  case,  may  still  be  found,  from  the 
same  theorems. 

Let  the  series  jn  +  el^  +  m  +  "2eY  +  m  +  ZeY  .  .  •  .  • 

m  -f  iiey  be  proposed,  where  m  and  e  denote  any  quan- 
tities whatever,  and  where  n  represents  the  number  of 
terms.  Then,  by  actually  raising  each  root  to  its  second 
power,  and  placing  the  terms  in  order,  the  given  expres- 
sion will  stand  thus : 

m^  -f    77i'  +    772^  ...  .  m^ 
2me  +  4me  +  6me  ....  2?ime  \^ .        Now,    it    is    evident 

e^    +  4^2  ^  9(?2 7i2 

that  the  sum  of  the  first  rank,  or  series,  is  7i  x  in^  -  also 
the  sum  of  the  second,  or  2me  X  1  -f  2  -f  3  -f  4  .  .  .  .  yz, 

Tl  X  71  ~  I    1 

appears  (<^^/  case  1)  to  be  2me  X ?   an(d  that  of 


'} 


the  third,  or  e^  x  1  +  4  -f  9  -f-  16 ti^  (^bi/  case  2) 

=  e^  X  — ^ i— :    therefore  the  sum  of  the 

6 

whole  progression,  m  +  eY  +  m  -{-  2ef  -f  m  +  SeY' 

•  .  .  .  w  -f  neY  is  =:  ?i  ,  7n^  +  n  .  ?i  +  1  >  me  -f 

n  .  71  -f  1  .  272  -f-  1  .  e^ 


6 

In  like  manner,  if  the  series  proposed  be 

m  +  eY  +  ~mr+2eY  +  w^  +  ^ef ^  +  ^^1^  J   then 

may  it  be  resolved 


into 


1  +  1-f-    1 1    X 


1+24-    3.  .  , 

.  .  n 

1  4-4-f    9.  .  , 

..7Z2 

.1 


X  ^nie 
.1+8+27 n^  X  ^    J 


X   3772^^   f  ,  1 

^  ^  :     whose  sum,  by 


the  Slims  of  Progressions.  SOT' 

the     Ibrementioned      theorems,      will      appear     to     be 
7z  .  /z  -f  1  •  Sm^e   .   n  .  7Z  -f  1  .  2/2  -f-  1  .  me^ 

«.  ^3  +    _ +   _ + 

!i_liLZ — !_£-,      And,  by  following  the  same  method, 

4  .        .  . 

the  sums  of  other  series  may  be  determined,  not  only 
of  powers,  but  likewise  of  rectangles,  and  solids,  fc^c. 
provided  their  sides,  or  factors,  be  in  arithmetical  pro- 
gression. Thus,  for  example,  let  there  be  proposed 
the  series  of  rectangles  m  +  e  .  p  -^  e  +  7?z  +  2^  .  p  +  2e 
-f-  m  '{'  oe  .  p  -{-  Se  .  •  ,  .  +  m  +  ne  .  p  +  ne.  Then,  the 
factors  being  actually  multiplied  together,  and  the  terms 
placed  in  order,  the  given  series  will  be  resolved  into  the 
three  following  ones : 

77ip     +      7np       +      mp       +       7Tip      . .  • .  +     inp 

m  -{-p.e  -\'m-{-p»2e  +  7n'{-p*3e  +771  -f-/?.4e'....-f-7/z-fj&.?2r 

ef2  +  4^2  +  9e^  +  1 6^2 f-         7i^e^. 

Whereof   the   respective    sums    {by  case  1   and  2)  are 

71  .  71  +  1  ,     „        n  ,  71  +  1  .  271  4^  1 

mp  X  n^Tu  +p  .  e  X 1 •,  and  e^  X ■ — : • — 

and  the  aggregate  of  all  these,  or 


n  +  1     .  7z  +  1  .  2?2  4-  1      ,    . 

71  X  mp  H —  .  m  +p  .  e  -J ■ ■ —  •  e^,  is  con- 

2  6 

sequently  the  true  sum  of  the  series  of  rectangles  pro- 
posed. 

From  this  last  general  expression,  the  number  of  can- 
non-shot in  an  oblong  pile,  whether  whole  or  broken, 
will  be  known.  For,  supposing  ^  =  1,  our  series  of 
rectangles  becomes  7n  +  1  .  p  +  1  +m+2.j&-t-2-f 
m  +  3  .  p  +  3 +  m  -\-7i  .  p  +  n  ;   and  the  sum 

thereof  =  n  X  mp  -\ -i-—  •  7n  +p  -j -I- -11.  = 

2  6 

the  number  sought:  where  m  +  1  and  p  +  ^  repre- 
sent ther  length  and  breadth  of  the  uppermost  rank,  or 
tire:    7i  being  the  number  of  ranks  one  above  another. 


208  The  Investigation  of 

But  the  expression  here  brought  out  may  be  reduced  to 


n  -f-  1   .72  —  1 


.--.  X  2/?2  +  7i  -f  1  *  2p  +  71  +  1  -{ — ',  which 

4  3 

is  better  adapted  to  practice,  and  which,  expressed  in  words^ 

gives  the  following  rule. 

To  twice  the  length,  and  to  twice  the  breadth  of 
the  uppermost  rank,  add  the  number  of  ranks  less  one, 
and  multiply  the  two  sums  together  ;  also  multiply  the 
number  of  ranks  less  one,  by  that  number  more  one, 
and  add  -^  of  this  product  to  the  former ;  then  -|  of  the 
sum  multiplied  by  the  number  of  ranks  will  be  the  an- 
swer. 

As  a  rule  of  this  sort  is  of  frequent  use  to  persons  con-^ 
cerned  in  artiller}',  it  may  not  be  improper  to  add  an  ex- 
ample or  two,  by  way  of  illustration. 

1.  Suppose  a  complete  pile,  consisting  of  15  tires,  or 
ranks,  and  suppose  the  number  of  shot  in  the  upper- 
most (which,  in  this  case,  is  a  single  row)  to  be  32* 
Then  the  first  product  mentioned  in  the  rule  will  be 
64  +  14  X  2  -f-  14  =  78  X  16  =  1248  ',  and  the  se- 
cond =  14  X  16  =  224;  \  whereof  is  r4|,  and  this, 
added  to  1248,  gives  1322|;  whereof  ^  part  is  330f  5 
which,  multiplied  by  15,  gives  4960,  for  the  whole  num- 
ber of  shot  in  such  a  pile. 

2.  Let  the  pile  be  a  broken  one,  such,  that  the  length 
and  breadth  of  the  uppermost  tire  may  be  25  and  16,  and 
the  number  of  tires  11. 

Here,  we  have  50 -f- 10  X  32  -f  10  =  60  X  42  =  2520j 
for  the  first  product ;  and  12  X  10  =  120,  for  the  second  : 

therefore  X  1 1  =  640  x  1 1  =  7040,  is  the  true  an- 

4 

swer. 

Having  exemplified  the  use  of  the  theorem,  for  find- 
ing the  sum  of  a  series  of  rectangles,  I  shall  here  subjoin 
one  instance  of  that  preceding  it,  for  determining  the 
sum  of  a  series  of  cubes  ;  wherein  the  value  of  the 
first  10  terms  of  the  progression  2  -f-  v  2^  +  3  +  2  V  ^^ 
4.  4  -f-  3  s/^y  +  5+4  \/TY  ^c.  is  required.     Here, 


the  Sums  of  Progressions,  209 

e  being  1  +  y/  2,  m  will  be  =  1  ;    therefore,  by  writing 
10,  1,  and  1  +\/2  for  7^,  w,  and  e^  respectively,  in  the 

Ml  1                .^        10.11.3.1  -f- ^2 
general  expression,  it  will  become  10  + — - 


10.  11  .21  .  1  +  V/^T  ,  100  .  121  .  1  +  V2]^  _ 
+  _  -  +  -  _ 

24815  +  ireOO  \/2,  the  value  sought. 

If  any  one  be  desirous  to  see  this  speculation  carried 
further,  so  as  to  extend  to  series  of  powers,  whose  in- 
dices are  fractions ;  such  as  square  roots,  cube  roots, 
&fc.  I  must  beg  leave  to  refer  to  my  Essays^  where  it  is 
treated  in  a  general  manner.  Here,  I  must  desire  the 
reader  to  observe,  once  for  all,  that  the  theorems  above 
found  will  hold  equally  true,  in  case  of  a  descending  series, 

such  as  m  —  ef  +  ^  "~-  ^^  j    ^^*  ^^  ^  —  ^1    +  ^'^ "~  ^^  I 
&Pc.  provided  the  signs  of  the  second,  fourth,  &fc.  terms 
be  changed  ;  as  is  evident  from  the  investigation. 

Although  the  subject  of  this  section  has,  already, 
been  pretty  largely  insisted  on,  yet  it  may  not  be  im- 
proper to  add  a  different  method,  whereby  the  same 
conclusions  will,  in  many  cases,  be  more  easily  derived : 
in  order  to  w^hic^  it  is  necessary  to  premise  the  subse- 
quent 

LEMMA. 

lfa  +  b  +  c  +  d  +  e  +  &fc.  be  a  series,  whereof 
the  terms,  g,  b^  c,  <^,  &:c.  are  so  related  to  each  other, 
that  the  sum,  or  value  thereof,  can  be  universally  ex- 
pounded by  an  expression  of  this  form,  viz.  An  -f  B  x 

^  ^  ^  — '  ^  +  ^  X  ^  X  n  —  1  X  n  —  2  +  D  X  ?i  X  72 1 

Xn  —  2  X  n  —  3  £sPc.  n  being  the  number  of  terms  to 
which  the  series  is  to  be  continued,  and  A,  B,  C,  D,  ^l\ 
determinate  coefficients  ;  then,  I  say,  the  values  of  those 
coefficients  will  be  as  hereunder  specified,  viz. 
A  =  a 

B=:-^  +  ^ 

2 
2E 


210  The  Investigation  of 

■""  2.3.4  ' 

p,  __a  —  4^  +  6c  —  4<d  +  c 
""  2.3.4.5  ' 

£sfc.  ^c. 


For,  since  the  equation  Ax^  +  Bx^^X^z  —  1  + 
C  X  n  X  n—-!  X  n  —  2  +  D  X  n  X  n  —  1  x  /z  —  2  X 
fi  —  3  £cf c.  ziza  +  b  +  c-i-d  +  e^  &c.  is  supposed  to  hold 
universally  true,  let  the  number  of  terms  be  what  it  will, 
let  n  be  expounded  by  1,  2,  3,  4,  cffc.  successively,  and 
the  general  equation  will  become 

^°.  2A  +    2B  =  «  +  <^, 
3°.  3A+    6B+    6C  =  a  +  /^  +  c, 
4^4A  +  12B+24C+    24D  z=a  +  b  +  c  +  d, 
5°.  5 A  +  20B  +  6OC  +  I20D  +  I20E  =za  +  b  +  c  +  d  +  e, 
&c.  &c. 

Now,  the  double  of  the  first  of  these  equations  being 
subtracted  from  the  second,  its  triple  from  the  third,  and 
its  quadruple  from  the  fourth,  &fc.  we  shall  have 
#2B  =  <^  — «, 

6B  +    6C  =  —  2a  +  b+c^ 
12B+24C+    24D=  — 3<x-f  ^  +  c+^, 
20B  +  6OC  +  I20D  +  1 20E  =  —  A.a  +  b  +  c  +  d'^-e^ 
&c.  &c. 

^  Again,  if  the  triple  of  the  first  of  these  be  subtracted 
from  the  second,  and  its  sextuple  from  the  third,  Ss^c.  we 
shall  next  have 

#6C  =  a  — 2<^+c, 

24C+    24D  =  3a  —  Sb-^-c-^d, 

6OC  +  I20D  +  I20E  =  e>a~9b  +  c  +  d+e. 

Moreover,  by  taking  the  quadruple  of  the  first  of  thes^ 
from  the  second,  £s?c.  we  get 

^^24D  +  —  «  +  3^  —  3c  +  ^,  and 

120D  +  120E=:— 4a  +  ll<^  — 9c  +  <^  +  ^j 


the  Swm  of  Progressions.  211 

from  the  latter  of  which  subtract  the  quintuple  of  the  for- 
mer, and  there  will  remain 

=^120E  =a  — 4^  +  6c  — 4^  +  ^. 
Now,  divide  each  of  the  equations  marked  thus  ('^),  by 
the  coefficient  of  its  first  term,  and  there  will  come  out 
the  very  values  of  A,  B,  C,  D,  csPc.  above  exhibited^ 
Q.  E.  D. 

COROLLARY. 

If  every  term  of  the  proposed  series,  a^  h^  c,  d^  8icc»  be 
subtracted  from  the  next  following,  the  first  of  the 
remainders,  —  a  +  ^,  —  ^  +  c->  —  c  +  ^y  —  ^  +  ^9  &c. 
divided  by  2,  gives  the  value  of  B,  the  coefficient  of 
the  second  term  of  the  assumed  series.  And  if  each  of 
the  quantities  thus  arising  be  subtracted  from  its  suc- 
ceeding one,  the  first  of  the  new  remainders,  a  —  2^  +  c, 
b  —  2c  +  d^  c  —  2d  +  e^  &c.  divided  by  6,  will  be 
equal  to  C,  the  coefficient  of  the  third  term  of  the 
same  series.  In  like  manner,  if  each  of  these  last  re- 
mainders be  again  subtracted  from  its  succeeding  one, 
the  next  remainders  will  be  —  a  +  Sb  —  3c  +  c/,  —  b 
+  3c  —  3d  +  e^  &:c.  whereof  the  first,  divided  by  24, 
gives  the  coefficient  of  the  fourth  term,  ££?c.  &fc.  There- 
fore, if  the  first  remainder  of  the  first  order  be  denoted 
by  P,  the  first  of  the  second  order  by  Q,  the  first  of  the 
third    by    R,    the  first    of   the    fourth  by  S,   ^c.   then 

^  being  =  B,  -^  =1  C, =  D,  ^ 


2  °  "2.3  '2.3.4  '243.4.5 

=  E,  &fc.  it  is  manifest  that  the  sum  of  the  series  a  +  b 
^.c  +  d+e-i-f    &c.    will    be    truly  expressed  by 

1 .  2 ^     __         1.2.3  ^ 

R  X  ^^^  —  ^  ^  ^  —  2  X7i  —  ^ 
1.2.3.4 


S  X  ^  ^  ^ —  lXy^~2x?^  —  3  X  n  —  4   np 
1.2.3.4.5  ' 

Example  1.     Let  the  sum  of   the  series  of   squares 
t  4.  4  -f-  9  +  16  4  .  .  *  +  71*  be  required;      Then,  ti«k- 


1   +  4  +  9  +  16  +  25   .  .  .  , 

>   • 

Sn  X  n — 1     J    n  X  n — Ixn- 

-2 

2     .        ■'                      3 

ZIa.  The  hvocsti^ation  of 

ing  the  differences  of  the  several  orders,  according  to  the 
preceding  corollary,  we  have 

1,4,9,  16,25,  36,  &c. 

3,5,    r,    9,  11,  &c, 

2,    2,    2,    2,  &c. 

G,    G,    G,  &c. 

Therefore  a  in  this  case  being  =  1,  P  =  3,  Q  =c  2, 

and  R,  S,  ££?c,  each  =  G,  the  sum  of  the  whole  series, 

w^,  is  found  z=z  n  + 
—  ^^^  +  3n^  +  n  _ 
~  6  ~ 

n  X  ^  +  1  X  2n  +  1 
6 

Example  2.  Let  it  be  required  to  find  the  sum  of  n 
terms  of  the  following  series  of  cubes,  viz.  27  +  64  + 
125  +  216  +  343  -f  512,  csfc.  Proceeding  here  as  in 
the  last  example,  we  have 

27,  64,  125,  216,  343,  512,  &c. 

37,    61,    91,  127,  169,  &c. 

24,    3G,    36,    42,  he. 

6,      6,      6,  &c. 

G,      O,  &c. 

Therefore,  by  substituting  27  for  a^  S7  for  P,  24  for  C, 

and  6  for  D,  we  thence  get 

^    37/z  X   71  —  1     .    24n  X  n  —  1   X   n  —  2     , 
27n  +  ~ .  + g-^ ~ + 


6^Xn-lX  ;z-2x^^-3      ^^j^.^^    abbreviated,  be- 
1.2.3.4  ' 

comes 1 1 1-  15;z,  the  sum,  or  value,  re- 

4^24^         ' 

quired  • 

Example  3.  Let  the  series  propounded  be  2  +  6  +  12 
+  2G  +  3G,  &fc.     In  this  case,  we  have 
2,  6,  12,  20,  30,  &c, 
4,    6,    8,  10,  &c. 
2,    2,    2,  &c. 


the  Sums  of  Progressions. 


213 


Hfence,  a  being  =  2,    P  =  4,    Q  =  2,   and  R,  S, 
^c.  each  =  0,  the  sum  of  the  series  will  therefore  be 

4.n  X  n  —  1       2nXn  —  1  Xn  —  2       ii^  +  3n^  +2n 
%n  + +  — ^ = T 


nxn-flx^i-f^ 


And,  in  the  same  manner,  the  sum  of  the  series  may  be 
truly  found,  in  all  cases  where  the  differences  of  any  or- 
der become  equal  among  themselves  :  and  even  in  other 
cases,  where  the  differences  do  not  terminate,  a  near  ap- 
proximation may  be  obtained,  by  carrying  on  the  process 
to  a  sufficient  length. 


SECTION  XV. 


Of  Figurate  Numbers^  their  sums^  and  the  sums  of 
their  reciprocals^  xvith  other  matters  of  the  like 
nature. 


2d- 
3d 

^ii1^th 

I  6th 

Lrthj 


1  •    1 

4.  5 
10 .  15 
20.  Z5 
35  .  70 


*     THAT    series    which    arises    by    adding    together  a 
rank 

'Units  (called  fig.  No.  of  the  1st  ord.) 

Figurate  numbers  of  the   2d  order 

p      Figurate  numbers  of  the  3d  order 

]  Figurate  numbers  of  the  4th  order 

!  Figurate  numbers  of  the  5th  order 

LFigurate  numbers  of  the  6th  order 

Therefore  the  figurate  numbers 

"Ist  order'l  f  1  .  1  .     1 

2d  order  1  |    1  .  2 .    3 

of  the  <  3d  order  ^  are  <{    1.3.    6 

4th  order  I  1  .  4  .  10 

L5thorderJ  Ll  .  5  .  15 

Hence  it  is  manifest,  that,  to  find  a  general  expression 

for  a  fig\irate  number  of  any  order,  is  the  same  thing 

as  to  find  the  sum  of  all  the   figurate  numbers  of  the 

preceding  order,  so  far.      Let  n  be  put  to  denote  the 


&c. 
&c. 
&c. 
&c. 
&c. 


214  The  Sums  of  Figurate  Numbers y 

distance  of  any  such  number  from  the  beginning  of  its 
respective  order,  or  the  number  of  terms  in  the  pre- 
ceding order  whereof  it  is  composed :  then  it  is  evident^ 
by  inspection,  that  the  sum  of  the  first  order,  or  the  ?7th 
term  of  the  second,  will  be  truly  expressed  by  72,  the 
number  of  terms  from  the  beginning.  It  is  also  evident, 
from  sect.  14,  p.  203,  that  the  sum  of  the  second  order, 

1+24-3+4 11,  will  be  !^  +  ii  (=  ii  X  !LdLi) 

^  2^2^12^ 

which,  according  to  the  preceding  observation,  is  also 
the  value  of  the  nth  term  of  the  third  order.  Hence, 
if  the  numbers,  1,2,  3,  4,  5,  fcPc.  be  successively  written 

instead  of  ;2,  in  the  general  expression 1 ,  we  shall 

thence  have  ^  +  i,|  +|,|  +|,  i_6  + 1^  2^^  + 1^  ^,^  foj. 
the  values  ot  the  first,  second,  third,  fourth,  fifth,  Sfc. 
terms  of  this  order,  respectively  ;  whence  it  appears,  that 
the  series  1+3  +  6+10  +  15  +  21,  &fc.  may  be  re- 
solved into  these  two  others,  viz. 

i +1  +  1 +l  +  V+¥,&c.  and 

l+l+.l+l+l  +  |,&c. 
The  former  of  which  being  a  series  of  squares,  its  sUm  ■will 

therefore  be  = 1 1 {by  case  2,  p.  203),  and 

that  of  the  latter  series  (<^2/  case  1,  p.  203)  appears  to  be 

1 :   and  the  aggregate  of  both,  which  is 

I 1 (or  —  X  X  )  will  be  the  true 

62^3^         1  2  3^ 

value  of  the  proposed  series  1  +  3+6  +  10  +  15 
l^c.  continued  to  ii  terms,  and  therefore  equal,  like- 
wise, to  the  72th  term  of  the  next  superior  order, 
1+4  +  10  +  20  +  35,  £ffc.  Let,  therefore,  1,  2, 
3,  4,  5,  &?c.  (as  above)  be  successively  written  for  72  in 

Q''^  7?^  72 

this  general  expression,  -^  -| +  "T^  ^^^  ^^  '^^  ^" 

come  1  4.  1  +1,  l+l  +1,  +  V  +1+1,  V  +  y  t> 
^c.  for  the  values  of  the  first,  second,  third,  fourth,  Isc, 


and  their  Reciprocals.  215 

terms  of  the  fourth  order,  respectively;  whence  it  ap- 
pears that  the  series  1  +  4  +  10  +  20  +  :^5^  ^c.  may  be 
resolved  into  these  three  others,  viz. 

1  +  8  +  2r  +  64  +  125+216  .  .  .  .  n^ 

^6  ' 

1  -f  4  +  9  +  16  +  25  +  36 n^ 

1  +2  +  3+4  +  5<^+  6 n  ^ 

3 

^       ,  72^  71^  71^     71^  71^  n 

whereof   the    sums    are    -^  +  ^  +  -^,  _  +  -_  +  _^^ 

and  —   +  ^  (by  p.  202    and  203)    the    aggregate  of 
6  6 

,  .   ,  n^  TV'  lln^      .      ^     /-  ^^  7Z    +    1 

Zl—  X  — -^^ — )  will  consequently  be  the  true  value  of 

3  4 

the  whole  series.     Aftet  the  same  manner  the  sum  of  the 

^  ^ ,  1  -n  I  ^^  72  +  1  72  +  2  71+5 

fifth  order  will  appear  to  be  —  X  — ^- —  X  —^ —  X 


n  +4 


12  3  4 

from  whence  the  law  of  continuation  is  ma- 
5  ' 
nifest.  And  it  may  not  be  amiss  to  observe  here,  that 
though  the  conclusions  thus  brought  out  are  derived  by 
means  of  the  sums  of  powers  determined  in  the  preced- 
ing section,  yet  the  same  values  may  be  otherwise  obtain- 
ed, by  a  direct  investigation,  from  either  of  the  two  gene- 
ral methods  there  laid  down. 

In  order  now  to  find  the  sum  of  the  reciprocals  of  any 
series  of  figurate  numbers ;  suppose  1  +  <^  +  ^c  +  bed 
+  hcde  +  hcdef  +  &c.  to  be  a  series  whose  terms  con- 
tinually decrease,  from  the  first  to  the  last,  so  that  the 
last  may  vanish,  or  become  indefinitely  small :  then,  by 
taking  the  excess  of  every  term  above  the  next  Ibllow- 
ing  one,  we  shall  have  1  —  h^h  X  1  —  c^  he  X  1  —  d^ 
bed  X  1  —  e^  hcde  x  1  —  /,  &c.     The  sum  of  all  which 


216  The  Sums  ofFtgurate  Numbers. 

is  evidently  equal  to  the  excess  of  the  first  term  above 
the  last,  or  equal  to  the  first  term,  barely;  because  the 
last  is  supposed  to  vanish,  or  to  be  indefinitely  small  in 
respect  of  the  first.  Hence  it  appears,  that  1  —  b  -{- 
b  X  1  -— c  +bc  X  1  — d  +  bed  X  1  —  e  +  bode  X  1  — /, 
&c.  =  1. 

Let  b  be  now  taken  =  -— ,  c  =  —^^  d  ■=.  — i-l 

a  a  +  p  ^  -f  ^  ' 

c  =  — lL_,  f  =  ,  &c.      Then,  1  —  b  being  = 

a  +  r  a  +  5 

a  —  m   ^               a  —  m             ,      a  —  m    .  a —  7n 

-,  1  —  c  =: ,  1  —  d= ,  1  —  e: 


a  a-j-p  a  +  q^  a -^  r^ 

he.  we  shall,  by  substituting  these  several  values  in  the 

,  a  —  ?n         m         a  —  m         m 

above  equation,  have  -| x 1 X 

^  a  a  a+p  a 

m  4-p      a  —  m   ^    m       vi  -{-  p      m  +  (/      a  —  w.o 

— 31^  X 1 X  ^  X  —^-^  X h  &c. 

a+p        ci  +  f]         ci        f^+/?        a  +  q        a  -^-r 

1                      1     ^     .       ^       .      w          ^  -{-  P  \ 
=  1  ;    and  consequently  1  ^ X + 

,..  X  — 3^  X  — -^^  +  &c.  =  ;   by  dividmg 

a+p        a  +  ^       a  +  r  a  —  m 

the  whole  by . 

a 

Hence,   if  q  be  taken  =  2/j,    r  ~   3p^  s  z=:  4/?,   &c. 

wi 
and  /3  be    put  =z   a  +  p^    we    shall    have    1    +  —    + 

m.m+p      7n . m+ p .  m  +  2p      m . m  +  p  .  m  +  2p  ..m  +  Sp 
jn+])  '^(i.^+p.^+2p'^(i.^+p.^  +  2p.i  +  3p 

J-  &c.  ad  infinitum^    =    ■ ;     which,    when 

^  ^  |3    p    771 

7)1       m  .  771  +  1       771 ,  7n  +  1  .  m  +  2, 
^  =  1,  becomes  1 +-+ ^-^- +  ^-^-j:-^-^^^ 

,    2^^^  ^  lZZ :     this,  by   taking  m  =  1,  and 

1  1.2  1.2.3 

s  =  n,  gives  1  H 1 "  -f       ==         ,  ^  "^ 


and  their  Reciprocals.  217 

2.3.4                  .    p            7^  —  1  , . 

+  6CC.  = ;    exhi- 


n  .  /i  4-  1  .  /z  4-2  .  n  -f-  3  n  —  2 

biting  the  general  value  of  a  series  of  the  reciprocals  of 
figurate  numbers,  infinitely  continued ;  whereof  the  or- 
der is  represented  by  n '  from  whence  as  many  parti- 
cular values  as  you  please  may  be  determined.  Thus, 
.  by  expounding  nhy  3,  4,  5,  £s?c.  successively,  it  appears 
that 

14 h  —  + f-  — ,  &c.  =  2, 

1  +  1  4.  i.  4. 1  +  1,  &c.  =  ^. 
^  4        10       20       35'  2 

14-  — +  —  +  —  +  — ,&c.  =  ±, 
5        15       35       70  3 

and  so  on,  for  any  higher  order  ;  but  the  sums  of  the  two 

first,  or  lowest  orders,  cannot  be  determined,  these  being 

infinite. 

By  interpreting  /3  and  7n  by  different  values,  the  sums 

of  various  other  series  may  be  deduced  from  the  same 

general    equation.       Thus,  in  the    first  place,    let  jS  = 

711 

m  4-  2  ;  so  shall  the  said  equation ,  become  1  4 ^  4- 

77Z  .  m  4-  1         ,        711  ,  m  4-  \         .        in  .  w  4-  1  « 

+  — — >  &c. 


w4-2.;?2  4-3       7^4-3.^72-1-4       /;z4-4.?/z4-5 
=  m  4-  1  ;  which,  divided  by  7?2  .  m  4-  1,  gives 

1.1.1.1 


m. 711  +  1      m  +  l  •711  +  2      Tn  +  2%7n  +  S      m  +  3 .7n+4> 

&c.  =  — . 
m 

Again,  by  taking-  3  2=  w  4-  3,  and  dividing  the  whole 
equation  by  771 .  m  +  1  •  ni  +  2^  we  have 

-__L__  + !_ — ,  + 

771  .  m  +  1  ,  m  +  2       7n  +  1  .  m  +k,  ,  m  +  3 


OT  +  2  .  w  +  3  .  OT  +  4.  m.?n  +  1  .2 

2  F 


2:18  The  Sums  of 

In  like  manner  we  shall  have  — = 


-,&c.=:.  ^ 


w -f- 1  .  w -4- 2  .  m -f  3  .  w +  4  m.m  +  \  .m  +  2  .^ 

From  whence  the  law  for  continuing  the  sums  of  these 
last  kinds  of  series  is  manifest ;  by  which  it  appears,  that 
if,  instead  of  the  last  factor,  in  the  denominator  of  the 
first  term,  the  excess  thereof  above  the  first  factor  be  sub- 
stituted, the  fraction  thence  arising  will  truly  express  the 
value  of  the  whole  infinite  series. 

A  few  other  particular  cases  will  further  show  the  use 
of  the  general  equations  above  exhibited. 

Let  the  sum  of  the  series   1+  —  J--1   x  —   + 

2  4         62         4         6         8o  ^  -   j:   -.         u 
—  X  —  X X  —  X  —  X  — n  Sec.  ad  znfimtum.  be 

5         7         9         5         7         9        11  -^ 

required. 

Here,  by  comparing  the  proposed  series  with  1  -^ 

-  P 

4-  =~-  +  &c.  (=  — - — ^ — ),  we  have  m  =  2, 

^.^+p  ^  —p  —  m 

3  =  5,  and  /?  =  2  ;  and  consequently  — —  =  3  = 

^         '  ^  ^         /   f3— /  —  m 

the  true  value  of  the  series. 

Let  the  sum  of  an  infinite  series  of  this  form,  viz. 

^ 1 —  -f  &c.  be  de- 


1  .  2  .  3,  &c.       2  .  3  .  4,  &c.       3  .  4  .  5,  &c. 
manded. 

Here    (according    to    the    preceding    rule)    we    have 

+    ^ o    +    t; :^    ^c-    =   -. :    =    ^  5 


.22.33.4'  1.1 

+  :; — : — r-)  ^c. 


1.2.32.3.4       3.4.5  1.2.2        4 

11  1__  ^^  ^  1  _   1, 

1.2. 3. 4*^2.3.4. 5      3.4.5.6'     ^'       1.2.3.3      is' 

£5fc.  i^c. 


Compound  Progressions.  219 

If,  instead  of  the  whole  infinite  series,  you  want  the 
sum  of  a  given  number  of  the  leading  terms  only  ;  then 
let  the  value  of  the  remaining  part  be  found,  as  above, 
and  subtracted  from  the  whole,  and  you  will  have  your 
desire. 

Thus,  for  instance,  let  it  be  required  to  find  the  sum 

of  the  ten  first  terms  of  the  series -\ ^ 

-f   ,  i^c.       Then  the  remaining  part, -f- 

1 1 . £s?c.  beincj  =  —  (hf  the 

12  .  13  ^  13  .  14  ^  14  -15'  ^        11  ^  ^ 

rule  above)  ^  and  the  whole  series  =1,  the  value  here 

souffht  will  therefore  be   1 =  — •       The  like  of 

^  11         11 

others. 

The  sums  of  series  arising  from  the  multiplication  of 
the  terms  of  a  rank  of  figurate  numbers  into  those  of  a 
decreasing  geometrical  progression,  are  deduced  in  the 
following  manner. 

By  the  theorem  for  involving  a  binomial  (given  at 
p.  40,    and  demonstrated   hereafter)    it    is  known    that 

n-  (or  1  —  .v]-'")  is  =  1  +  mx  +  m  •  ^^i^li  .  x"" 

1  —  X I  2  ^ 

m  +  1     m  +  2       „    .  m  +  1     711-^2    m  + 3 

^23  ^234^ 

&c.     In  which  equation  let  in  be  expounded  by  1 ,  2,  3,  4, 

5,  &c.  successively,  so  shall 

1°.  =  1   +  ;^  +  :^2  ^  -^^,3  ^  ^,4  ^  ^5  ^  ^^^ 

1  X 

2^         ^        =  1  +  2a;  +  3x^  +  4a,3  +  5.v^  +  6.r%  &c« 
1  —  x\ 

3°. =  1  +  3.V  +  Q>x^  +  10;^"'  +  15x^+  21a:%  &(^. 

1  —  x\ 

4^°-  =  1  +  4.V  +10v-  +  20a-  4-  o5.v^  4-  56.^^^c; 


226  The  Sums  of 

5\        ^     ,  =  1  +5Ar  +  15x^  +  35x^  +  YOx^  +  126;^^Scc* 
1  — x\ 

6^  ,  '  '•'  '.  =  1  +  6;tf  +  21.^2  +  5&x^  + 126;^'*  +  252;^*, &c» 
1  —  ;c  I 
All  which  series  (whereof  the  sums  are  thus  given)  are 
ranks  of  the  different  orders  of  figurate  numbers,  multi- 
plied by  the  terms  of  the  geometrical  progression  1,  x,  x^^ 
a:',  x*^  &c. 

From  these  equations  the  sums  of  series  composed 
of  the  terms  of  a  rank  of  powers,  drawn  into  those  of 
a  geometrical  progression,  such  as  1  +  4x  +  9x^  -f  16;c^, 
&€•  and  1  +  8x  -j-  27x^  -f  64^^,  &c.  may  also  be  de- 
rived ;  there  being,  as  appears  from  the  former  part  of 
this  section,  a  certain  relation  between  the  terms  of  a 
series  of  powers  and  those  of  figurate  numbers  ;  the  lat- 
ter being  there  determined  by  means  of  the  former.  To 
find  here  the  converse  relation,  or  to  determine  the 
former  from  the  latter,  it  will  be  expedient  to  multiply 
the  several  equations  above  brought  out,  by  a  certain  num- 
ber of  terms  of  an  assumed  series  1  -f-  Ax  +  Bx^  -|-  Cx^^ 
&c.  in  order  that  the  coefficients  of  the  powers  of  x  may, 
by  regulating  the  values  A,  B,  C,  D,  &^c.  become  the  same 
as  in  the  series  given. 

Thus,  if  the  series  given  be  1  -f  4x  +  9x^  +  16.^^  -f 

25a;'*,  &c.  ;     then,    by    multiplying    our    third    equation 

1  -f-  Aa:  

by  1  +  Ax,  we  shall  have  =====^  =z  1  -{•  3  +  A  X  x^ 

1  • — '  X  j 

+  6  +  3A  X  x^  4-  10  -f-  6 A  X  x^  +  &c.  which  series,  it 
is  tivident,  by  inspection,  will  be  exactly  the  same,  in  every 
term,  with  the  proposed  one,  if  the  quantity  A  be  taken 
=  1.     The  sum  of  the  said  series,  infinitely  continued,  is 

1  -4-  iV 

therefore  truly  represented  by       ~     -, 

-f 

In  lite  manner,   if  the  fourth  equation 


1  +  4.V  +  10^'  +  20.v3  ^  g5_^4^  gjc.  be  multiplied  by 


Compound  Progressio7is»  221 

A       .    T>  9     u           Ml      •      1  4-  A^c-  4-  Bx2        ^     , 
1  +  A.r  +  B:v^5  there  will  arise  — — —   =  1  + 

1  —  x\ 

4-f  A  X  ^  +  10 -f.  4A  -f-B  X  ^^  +  20 +  10 A  -f  4B  X  x^j 
&c.  where,  the  several  terms  of  the  series  being  com- 
pared with  those  of  the  series  1  +  8.r  +  27 x^  +  64.\^,  &c. 
we  have  4  4-  A  =  8,  and  10  4-  4A  4-  B  =  27  ;  whence 
A  =  4,  and  B  =  1  ;    and  consequently,  by  substituting 

these  values,  ■■     .     ■  -— -  =  1  +  8jc  4-  27^  4-  64^^  + 

i  —  xy 

125jv^&c. 

Again,  by  multiplying  the  fifth  equation,  = 

1  —  x^ 
\^5x  +  15 x^  4-  35;c3,  &c.  by  1   +  A^t-  4-  B;^^  ^  ^^3^ 

.    -  1  4-  A;c  4-  B;c2  4-  Cx^        ^         ~ 

it  becomes    — i ■ =  14-54-Axa;'  + 

1  —  x] 

15  4.5A  +B  X  ^^  +  ^5  +  15A  4.  5B  4-  C  X  a:^  &c. 
And,  by  comparing  the  several  terms  of  the  series  with 
those  of  1  4.  16x  +  SlJt^  4.  25^x^,  &c.  we  get  5  4-  A 
=  16,  15  4-  5A  4-  B  =  81,  and  35  4-  15A  4-  5B  4- 
C  =  256:  whence  A  =  11,  B  (=81  —  15  —  55)  =  11, 
and  C  (  =  256  —  35  —  220)  =  1  ;  and  consequently 
1  4.  Hat  4- 11.^^ -f  .Y^ 

--I —  ■ =  1  4-  16^  4.  Ux^  +  256x\  &c. 

1  —  xy 

By  proceeding  the  same  way,  it  will  be  found,  that 
1  4-  26x  4-  66;c2  4-  v6;c^  -h^^^       . 

-^ 7; =^^—  =  1  +2^^  +  3V  4.  4^Ar^  + 

1  —  X  \ 

&c.  &c. 

And,  itniversally^  putting  a  •=.  vi^  h  ^=:.  m  *  ^  "^    .^ 

m  4-  1      7?i  4-  2    p  J        1  .  ,  . 

ci=zm^  — - —   .  - — - — ,  &c.  and  multiplying  the  gene- 

ral  equation  --  —  1  +  ax  +  bx^  +  cx^  4.  dx\  &c. 

1— ;vT"  -r  -r         , 

by  1   4-  A.Y  4-  Bx^  4.  C;v3   ^  j^^^    g^^^  ^^^j.^  ^^.jg^^ 

1    4-    Ay    -L    Ba:#,    &c. 

"  T^^^T^ =l4-«-f-Ax^    + 


^22  The  Sums  of 

b  +  aA  +  B  X  ^^  +  c  +  bA  +  aB  +  C  X  oc^^  &<i. 
The  terms  of  which  series  being  compared  with  those 
of  the  series  1  +  2'"x  +  S^a;^  +  4Px^  +  5":v^,  &c. 
we  have  A  =  2"  —  a,  B  =  3"  —  a  A  —  ^,  C  ==  4«  — ^ 
^B  —  Z'A  —  c,  D  =  5^^  —  aC  _  ^B  —  cA  —  ^,  &c. 
where  the  law  of  continuation  is  manifest ;  and  where, 
from  the  law  observed  in  all  the  preceding  cases,  it  ap- 
pears, that  the  value  of  m  must  exceed  the  index  n^  of 
the  given  series  of  powers,  by  a  unit ;  and  that 
the  series  1  +  Ax  +  hx^  +  C:^^^,  &c.  will  always  con- 
sist of  n  terms ;  whereof  the  coefficients  of  the  first  and 
last,  the  second  and  last  but  one,  £5?c.  will  be  respectively 
equal  to  each  other:  so  that,  having  found  from  the 
preceding  equations  as  many  of  the  quantities  A,  B,  C, 
&Pc.  as  are  expressed  by  ^ii  —  1,  the  others  will  be  given 

^  ,  ,  ,1-1.  Ax  4-  Bx^  4-  Cx^^  &c. 
trom  thence,  and,  consequently, .       ^^ 

the  true  value  of  the  proposed  series  1   +  2";^  +  ^"-^^ 
-}-  4Px^^    &c.       Thus,    for  example,  let  tz  =  6 :     then 
?w  =  7  =  G,  /^  =  28,  A  =  64  —  7  =  5r,  B  =  729  — 
399  —  28  =  302  ;  and  therefore 
1+57X  +  302^^  4-  302x^  +  57x^  +  x'  =  1  +  2^x  4-  S^x^ 

-f-  4^x'^,  &c.  and  so  of  others. 

These  equations,  or  theorems,  give  the  sutn  of  the 
whole  series,  infinitely  continued ;  but  from  thence 
the  sum  of  any  assigned  number  of  terms  may  be  deter- 
mined, not  only  when  the  coefficients  are  a  series  of 
powers,  but  likewise  when  they  are  produced  by  factors 
that  are  unequal :  the  method  of  which  I  shall  in- 
stance in  finding  the  sum  of  t  terms  of  the  series 
f—P^g~9  •  z*-  +7-^  .^  —  27  .zr-\-'o  +f—3p. 
g  —  3^  .  zr+2v  -f.  Sec.  Which  series,  by  actually  multi- 
plying the  factors  together,  is  resolved  into  the  three  fol- 
lowing ones. 


Compound  Progressions*  223 


^fq+gp  .  2^  X  1  +  22^  -4-  32^^  -f  42^^  &c. 

+  /?y2^"  X  1  +  4^2"  +  ^^^^'^  +162^^  &c. 

The    sum    of   the    first    of    these,    infinitely    continued, 

supposing  X  =  2^,  will  be  =  -^ ;    that  of  the  se- 


cond =  —  y<7   -f  Rp  >  z_  .     ^^^  ^l^g^^  ^£  ^^    ^^^^  _ 
1  —  x]2 

— -^- *-l3—  ^  by  ivhat  has  betn  above  determined ;    and 

1  —  xY 

consequently     the     sum     of    all     the     three     equal     to 
_£_  ^f^^fj+lP  +  Pl_:J^  ^  ,he  whole  infi. 

nite  series  y —  Z'  •  ^  —  q  *^^  +f —  ^P  *  g  —  2^  •  2''+^  +v 
&Pc.  But  the  sum  of  the  t  first  terms  only  is  wanted ; 
therefore  the  sum  of  all  the  remaining  terms,  after  the 
t  first,  must  be  found  in  like  manner,  and  be  deducted 
from  the  sum  of  the  whole,  here  given.  Now,  to  do 
this,  we  are  first  to  get  the  leading  term  of  the  said  re- 
maining ones  I  which,  according  to  the  law  of  the  series, 
will  be  expressed  by  f —  p  —  tp  .  g  —  q  —  tq  .  2'+^  : 
whence,  if  we  make  f  —  tp  z=i  h^  g  —  tq  •=.  k^  and 
r  +  tv  :=.  s^  It  is  evident,  that  the  series  to  be  deducted 
will  be  h — p  .  k  —  q  •  2^  +  A  —  2/?  .  k  —  2q.  2*+^,  &c. 
which,  having  the  very  same  form  with  that  first  pro- 
posed, its  sum  will  therefore  be  had  by  barely  writing 
h  for  f^  k  for  g^  and  s  for  r,  in  the  value  above  deter- 
mined :  which,  thereby,  becomes 


224  The  Sums,  ^e. 

In  the  same  wanner,  supposing  the  t  first  terms  of  the 
series  a  —  .  d  — p  ,  c  — /? .  d — p  .  &c.  x  z**  +  a  —  ^p  . 
Ij  —  zp  .  c — ..p  .  d —  -/?  .  &c.  X  z^'+S  &c.  were  to 
be  required ;  by  putting  the  continual  product  of  all  the 
quantities  a,  h,  c,  d,  &c.  =  P ;     the    sum    of    all    the 

P  P  P 

products  ( f-  --   ^ ,  &c.)  that  arise  by  omitting 

one    letter     in     each,    =    Q;      the    sum    of   all    those 

P  P 

(-7    +   —1   &^c.),    by  omitting  two    letters,    =   R,   £sr*c. 

we  shall  here  have 


,s 


L       1  —  ^1  1  —  ^1 

&c.  for  the  sum  of  the  whole   infinite   series :     and  if 


/ 


we  make  «  =  a  -^  tp,  b  r=z  b  —  tp^  r  =z  r  +  tv^  &c. 
it  is  evident  that  the  sum  of  the  remaining  terms,  after  the 
t  first,  will  be  truly  expressed  by 


&c.  where  x  =  2«,  and  P,  Q,  R,  S,  cs^c.  are  the  same  in 

/     /     /     y 

relation  to  a^  Z>,  c,  ^,  &c.  as  P,  Q,  R,  S,  £^c.  in  respect  to 
a  b^  c,  d^  &€• 

A  multitude  of  other  cases  and  examples  might  be 
given,  there  not  being,  in  the  whole  scope  of  the 
mathematical  sciences,  a  subject  of  greater  variety  and 
intricacy  than  this  business  of  series  :  but  to  pursue  it 
farther  here  would  be  inconsistent  with  the  general  plan 
of  this  work.  Such,  therefore,  as  are  desirous  of 
a  greater  insight  into  the  matter,  may,  if  they  please, 
turn  to  my  Miscellanies^  where  it  is  carried  to  a  greater 
length. 


Of  Combinations »  22S' 

,  From  the  series  for  figurate  numbers,  derived  in 
the  former  part  of  this  section,  the  investigation  of 
a  general  theorem  for  determining  how  many  dif- 
ferent combinations  any  number  of  things  will  admit 
of,  when  taken  two  by  two,  three  by  three,  &fc,  may  be 
very  easily  deduced.  Let  the  number  of  things  in  each 
combination  be^  jirst^  supposed  txvo  only ;  and  let  n  hcy 
universally^  put  to  represent  the  xvhole  number  of  things  or 
letters^  «,  ^,  c,  d^  &c.  to  be  combined.  When  the  num- 
ber of  things  is  only  two,  as  a  and  b^  it  is  evident  that 
there  can  be  only  one  combination  («/?)  ;  but,  if  n  be 
increased  by  1,  or  the  letters  to  be  combined  be  three, 
as  a,  ^,  c,  then  it  is  plain  that  the  number  of  combina- 
tions will  be  increased  by  2,  the  number  of  the  preced- 
ing letters  a  and  b  ;  since,  with  each  of  those,  the  new 
letter  c  may  be  joined  ;  and  therefore  the  whole  num- 
ber of  combinations,  in  this  case,  will  be  truly  ex- 
pressed by  1  -f-  2.  Again,  if  n  be  increased  by  one 
more,  or  the  whole  number  of  letters  be  four,  as  a,  b^ 
o,  d;  then  it  will  appear  that  the  number  of  combina- 
tions must  be  increased  by  3,  since  3  is  the  number  of 
the  preceding  letters,  with  which  the  new  letter  d  can 
be  combined,  and  therefore  will  here  be  truly  ex- 
pounded, by  1  -f-  2  -f-  3.  And,  by  reasoning  in  the 
same  manner,  it  will  appear,  that  the  whole  number  of 
combinations  of  two,  in  five  things,  will  be  1  -f-  2  -f 
3  -f  4 ;  in  six  things,  l+2  +  3-f-4-|-5;  and  in 
seven,  l-f2  +  3-f-4-f5-f6,  &Pc.  Whence,  uni- 
versally, the  number  of  combinations  of  ?z  things,  taken 
two  by  two,  is  =  1  +  2  -}-  3  -f  4  -f-  .  .  •  ,  .  n  —  1  : 
which  being  a  series  of  figurate  numbers  of  the  second 
order,  where  the  number  of  terms  is  n  —  1,  the  sum 
thereof,  by  case  1,  p.  203,  will  tlierefore  be  truly  defined 
.     n  —  1        n  n  —  1 

Let  210ZV  the  iiumber  of  quantities  iri  each  covibiiiation  be 
supposed  to  be  three. 

Tf-   h  plain,  that,    in  three  things,   rv,   b,   (\  there   rrm 
^  G 


226  Of  Combinations. 

l)e  only  one  combination  ;  but,  if  n  be  increased  by  1, 
or  the  number  of  things  be  4,  as  a,  b^  c,  d^  then  will  the 
number  of  combinations  be  increased  by  (3)  the  number 
of  all  the  combinations  of  two,  in  the  preceding  letters 
rt,  3,  c ;  since  with  each  two  of  those  the  new  letter  d 
may  be  combined ;  therefore  the  number  of  combina- 
tions, ia  this  case,  is  1  +  3.  Again,  if  n  be  supposed 
to  be  increased  by  1  more,  or  the  number  of  letters  to 
become  five,  as  a,  ^,  c,  d^  e ;  then  the  number  of 
combinations  will  be  increased  by  six  more  (=14-2 
4-  3),  that  is,  by  all  the  combinations  of  two,  in  the 
four  preceding  letters,  «,  b^  c^  d:  since,  as  before,  with 
each  two  of  those,  the  new  letter  e  may  be  combined; 
Hence  the  number  of  combinations  of  n  things,  taken 
three  by  three,  appears  to  be  1  +  3  -f-  6  +  10,  &c. 
continued  be  ?2  —  2  terms  ;  which  being  a  series  of  figurate 
numbers  of  the  third  order,  the  value  thereof,  by  what 
is  before  determined  (p.  214)  will  be  tiTily  expressed  by 

n  —  2      71  ~  1        n  .  ^    n       n  —  1       n  —  2 

X X  — ,  or  Its  equal,  —  X X • 

1  2  3  ^      '  1  2  3 

And  universally,  since  it  appears  that  increasing  the 
number  of  letters  by  1  always  increases  the  number 
of  combinations  by  all  the  combinations  of  the  next  in- 
ferior order  with  the  preceding  letters  (for  this  obvious 
reason,  that  to  each  of  these  last  combinations,  the  new 
letter  may  be  joined),  it  is  manifest,  that  the  combina- 
tions, of  any  order,  observe  the  same  law,  and  are  ge- 
nerated in  the  very  same  manner  as  figurate  numbers, 
and  therefore  may  be  exhibited  by  the  same  general 
expressions  ;  only,  as  there  are  2,  3,  4,  5,  £s?c.  things 
necessary  to  form  the  first,  or  one  single  combination, 
according  to  the  different  cases,  it  is  plain  that  the 
number  of  terms  must  be  less  by  1,  2,  3,  £sPc.  respec- 
lively,  than  (n)  the  number  of  things ;  and,  therefore,^ 
instead  of  7Z,  in  the  aforesaid  general  expressions,  we 
must  substitute  n-. —  1,  ?2  —  2,  or  ;z  —  3,  £s?c.  respectively, 
to  have  the  true  value  here.  Hence,  the  number  of 
combinations  of  two  things,  in  n  things,  will  be 
a  —  1        n  n       n  —  1       n.         n  —  2       n — 1^     ^2 

— -  ^  -TT'  ^^'  Y  ^  --^ '  of  three,  -.-^  x  —^  X  y. 


Of  the  Binomial  Theorem.  227 

n       7^  —  1      n  —  2       r,r,        n  —  3      n  —  2      n  —  1 
«r_X-^  X  -^;  offour,__x-^X  -^ 

n  n       n  —  1      n  —  2      n  —  3  ,  .  ,         ^^  ^^ 

X  — ,  or  —  X X   — :; —  X   —   (vid.  p.  215)  : 

41  2  3  4^^^ 

whence,  universally,  the  number  of  combinations  in  the 

number,  tz,  of  things,  taken  two  by  two,  three  by  three, 

cSTc.  will  be  expressed  by  —  X X  — ; —  X , 

^c.  continued  to  as  many  factoids  as  there  are  things  in 
each  combination. 

From  this  last  general  expression,  showing  the  com- 
binations which  any  number  of  quantities  will  admit  of, 
the  known  theorem  for  raising  a  binomial  to  any  given 
power,  is  very  easily  and  naturally  derived. 

For  it  is  plain  that  <^«  [T     >a  +  bc=.a-\'bxa  +  C', 

which,  multiplied  by  a  +  d^  gives  a^  +  cV  a^  ^  bdy  a  + 

O  cd) 


+  d] 


bcdz=.  a  +  b  X  a-\-c  X  a  -{-d;  and  this,  again,  multiplied 
by  a  +  e,  gives 

+  b^ 
+  c 


^'++d 
+  e. 


+  dej 


+  bd\  +  brd^ 

+  c^  I  +  cdej 


a  +  bXa-^-cxa  +  dxa  +  e. 

Whence  it  appears,  that  the  coefficient  of  a,  in  the 
second  term,  is  always  the  sum  of  all  the  other  quantities 
^,  c,  J,  &c.  added  together ;  and  that  the  coefficient  of 
the  third  term  is  the  sum  of  all  the  products  of  those 
quantities,  or  of  all  their  possible  combinations,  taken 
two  by  two ;  since,  from  the  nature  of  multiplication, 
they  must  be  all  concerned  alike,  in  every  term  :  whence 
it  is  also  manifest,  that  the  coefficient  of  the  fourth  term 
must  be  the  sum  of  all  the  solids  of  the  same  quantities,  or 
of  all  their  possible  combinations,  taken  three  by  three, 


228  Of  the  Binomial  Theorem. 

Hence,  if  the  number  of  the  quantities  b^  c,  d^  <?,  &c. 
or  the  number  of  the  factors,  a  +  b^  a  +  c^  a  +  dy 
a  •}-  e^  &c.  to  be  muhiplied  continually  together,  be 
denoted  by  7z,  it  follows,  that  the  number  of  letters,  or 
quantities,  in  the  coefficient  of  the  second  term  of  the 
product  will  likewise  be  denoted  by  ?i ;  that  the  num- 
ber of  all  their  products,  or  of  all  the  combinations  of 

ji 1 

two,  in  the  coefficient  of  the  third  term,  will  be  7i  x  

2 

(it  having  been  shown  above  that  the  number  of  com- 

^i, 1 

binations  of  ?2  things,  taken  two  by  two,  is  tz  x  )  ; 

and  that  the  number  of  all  the  solids  of  those  quantities, 
or  all  the  combinations  of  three,  in  the  coefficient  of  the 

71  '        1         71  — —  2 

fourth  term,  will  ht  7i  X X  ,  &Pc.    Therefore, 

if  all  the  quantities  ^,  c,  d^  e^  &c.  be  now  taken  equal 
to  each  other,  it  is  evident,  that  a  -{-  b  x  a  -^  c  X  a  +  d 
X  CL  -\-  e^  &c.  will  become  a-^-bxa-^-bx^-^bxci  +  b^ 
&c.  or  a  4-  d/"]^^  and  that  the  coefficient  of  the  power 
of  <7,  in  the  second  term  of  the  product,  will  be  7ib  ;  in  the 

n 1 

third  71  X  — I — -    h'^  (since  all  the  rectangles,  as  well  as 

all  the  solids,  &?c.  do  here  become  equal)  ;    and  in  the 

Yi  \  yi  -        2  * 

fourth  n  X X -b^.  Sec.    But  it  is  evident,  from 

2  3  ' 

the  nature  of  multiplication,  that  the   powers  of   a^  in 

the  second,  third,  fourth,  £5?c.  terms  of  ^  +  Z?  raised  to 

the    powder    7?-,    are    fl;«-i,   a'^-'^^    a"-^,   £s?c.         Therefore 

a  +  />«,  ox  a  •\-  b  raised  to  the  power  7z,  is  truly  ex- 

71  '         1  71  '         1 

pressed  by  a^  +  7iba^~'^  +  7iX b^a^"^  +  w  X  

X  ^^~'    b^  ci""-^-,  i^c.  or  a^  +  72rt«~^  b  +  ?i  X  ^^  "~     a'^'-^S^ 

72  — —  1  71        -  2 

+  71  X  — - —  X  — :; —  a^'^^b^^  ^c.  as  was  to  be  shown. 


Of  Interest  ayid  Ammities.  ^29 

SECTION  XVL 

Of  Interest  and  Annuities. 

INTEREST  may  be  either  simple  or  compound: 
simple  interest  is  that  which  is  paid  for  the  loan  of 
any  principal,  or  sum  of  money,  lent  out  for  some  li- 
mited time,  at  a  certain  rate  per  cent,  agreed  upon  be- 
tween the  borrower  and  the  lender,  and  is  always  propor- 
tional to  the  time.  Thus,  if  the  rate  agreed  upon  be  4 
per  cent,  per  ann.  or,  which  is  the  same  thing,  if  the  interest 
of  100/.  for  one  year  be  4/.  then  the  simple  interest  of  the 
same  sum  for  two  years  will  be  8/. ;  for  three  years  12/. ; 
and  for  four  years  16/. ;  and  so  on  for  any  other  time  in 
proportion. 

Compound  interest  is  that  which  arises  by  leaving  the 
simple  interest  of  any  sum  of  money,  after  it  becomes 
due,  together  with  the  principal,  in  the  hands  of  the 
borrower,  and  thereby  converting  the  whole  into  a  ne^7 
principal.  Thus,  he  who  lets  out  100/.  for  one  year, 
at  the  rate  of  4  per  cent,  has  a  right  to  receive  104/.  at 
the  year's  end ;  which  sum  he  may  leave  in  the  borr 
rower's  hands,  a  second  year,  as  a  new  principal,  in 
order  to  receive  interest  for  the  whole  ;  and  this  inte- 
rest (which  will  be  found  4/.  Ss.  2|^.),  together  with 
4/.  the  interest  of  the  first  principal,  for  the  first  year, 
will  be  the  compound  interest  of  100/.  for  two  years  ; 
and  so  on,  for  any  greater  number  of  years.  But  I  shall 
first  give  the  investigation  of  the  theorems  for  simple 
interest. 

Let  the  rate  per  cent,  or  the  interest  of  100/.  for  one 
year  =  r  ;  the  months,  weeks,  or  days  in  one  year  =  t ; 
the  months,  weeks,  or  days  which  any  sum,  a^  is  lent  out 
for  =  n ;  and  the  amount  of  that  sum,  in  the  said  time, 
viz.  principal  and  interest,  =  b. 

Then  it  will  be,  as  100  is  to  r  (the  interest  of  100/.)  so 

is  the  proposed  sum  (a)  to  — ,  the  interest  of  that  sum, 
for  the  same  time.        Again,  as  f,  the  time  in  which 


230  Qf  Interest  and  Animifies* 

the  said  interest  is  produced,  is  to  n  (the  time  proposed),  so 

is ,  the  interest  in  the  former  of  these  times,  to  • 

100  '        100^* 

that  in  the  latter ;  which,  added  to  «,  the  principal,  gives 

a  -I =  b.  the  whole  amount:     from   whence  we 

100^  

-      ,  100^?  100?  X  b  —  a        J 

also  have  a  = ,  r  =  and  n  =: 

100?  +  nr  an  ' 

:  the  use  of  which  equations,  or  theorems, 

ar 

will  appear  by  the  following  examples  : 

Examp,  1.  What  is  the  amount  of  550/.  at  4  per  cent. 
in  seven  months  ? 

In  this  case  we  have  a  =  550,  r  =  4,  ?  =  12,  7z  =  7 : 

,  ^1         .    ^^^        ^^^   .  550X7X4  J 

and  consequently  a  -j =  550  -4 =  562|-/. 

^         ^  toot  100  X  12  ^ 

or  562/.  16^.  8d,  the  true  value  sought. 

Examp.  2.  What  is  the  interest  of  1/.  for  one  day,  at 
the  rate  of  5  per  cent.  ? 

Here  r  being  z=  5^  t  =  365,  a  =  1,  and  /z  =  1,  we  have 

^IlL  = £ =  1 =  0.0001369863,  £s?c.  = 

100?       100  X  365        100  X  73 

the  decimal  parts  of  a  pound  required. 

Examp.  3.  What  sum,  in  ready  money,  is  equivalent 
to  600/.  due  9  months  hence,  allowing  5  per  cent,  dis- 
count ? 

Here  r  being  =:  5,  ?  =  12, 72  =  9,  and  b  =  600,  we  have 

a  (by  theorem  2)  =  i22iL^22iLil  =  578,313/.  or  578/.. 

^^  ^       100X12  +  9X5 

6s.  3\d,  which  is  the  value  required. 

Examp.  4.  At  what  rate  of  interest  will  300/.  in  fifteen 
months  amount  to,  or  raise  a  stock  of,  330/.  ? 

In  this  case,  we  have  given  ?=12,  n  =15,  a  =  300, 
and  ^  =  330 ;    whence  \by  theorem  3)  r  will  come  out 

100  y  12  V  30 

= — — =  8  ;  therefore  8  per  cent,  is  the  rate 

300  X  15  ^ 

required. 


Of  Interest  and  Annuities.  231 

Examp*  5.  In  how  many  days  will  365/.  at  the  rate  of 

4!  per  cent,  amount  to,  or  raise  a  stock  of,  400/.  t 

ri       1  .N  u  100  X  365  X  ^^ 

Here  (by  theorem  4)  we  nave  n  =  ^ . 

^  ^  ^  365  X  4 

:=5  ^7S  =  the  number  of  days  required. 


Of  Annuities  or  Pensions  in  Arrear^  computed  at 
Simple  Interest. 

Annuities  or  pensions  in  arrear  are  such  as,  being- 
payable,  or  becoming  due,  yearly,  remain  unpaid  any 
number  of  years  :  and  we  are  to  compute  what  all  those 
payments  will  amount  to,  allowing  simple  interest  for  their 
forbearance,  from  the  time  each  particular  payment  be- 
comes due  :  in  order  to  which, 

TA  =.the  annuity,  pension,  or  yearly  rent. 
Y       J   7Z  =  the  time,  or  number  of  years,  it  is  forborne. 

j    r  =  the  interest  of  1/.  for  one  year. 

L.  7n  =  the  amount  of  the  annuity  and  its  interest. 
Then,  as  1  :  r  :  :  A  :  r  A,  the  interest  of  the  proposed 
sum  or  pension  A,  for  one  year  ;  which,  as  the  last  y earl's 
rent  but  one  is  forborne  only  one  year,  will  express  the 
-whole  interest  of  that  rent,  or  payment :  moreover,  since 
the  last  year's  rent  but  two  is  forborne  two  years,  its  in- 
terest will  be  2r  A  :  and,  in  the  same  manner,  that  of 
the  last  year's  rent  but  three,  will  appear  to  be  3rA,  ^c. 
&?c.  whence  it  is  manifest  that  the  sum  total  of  all  these, 
or  the  whole  interest  to  be  received  at  the  expiration  of 
n  years,  for  the  forbearance  of  the  proposed  annuity  or 
pension,  will  be  truly  defined  by  the  arithmetical  pro- 
gression r A  +  2r A  +  3r A  +  4r A  +  5rA,  fcfc.  continued 
to  n  —  1  terms,  that  is,  to  as  many  terms  as  there  are 
years,    excepting  the  last.       But  the  sum  of  this  pro- 

n  -—  1 
gression  is  equal  to  n  X X  rA  {by  theor.4,sect.  10). 

Therefore,  if  to  this  the  aggregate  of  all  the  rents,  or  ?zA, 
be  added,  we  shall  have  nA  -f '^^^^    X  ^A  =  m  : 


232  Of  Inter ent  and  AnnuUie^^ 


whence  we  also  have  A  =  n  —  1  ,  r  =^ 

n  4-  n  X  X  r 

2 

2wz  — 27iA  ,  J2m  1  .         . 

== ,  and  72  r=  W~~  +  />2  — p  .^  supposmg  /' 

nxn  —  ixA  ^  rA 

_  1  _  1 

""7"      '2" 

Examp*  !•  If  600/.  yearly  rent,  or  pension,  be  for- 
borne five  years,  what  will  it  amount  to,  allowing  4  per 
cent,  interest  for  each  payment  from  the  time  it  becomes 
due? 

Here  we  have  given  A  =  600,  n  =  5,  and  r  =  .04 
(for  as  100  :  4  :  :  1  :  .04)^  which  values  substituted,  in 

theorem  1,  give  m  =  (72 A  •\-  n  x  Ar  =  3000  + 

240)  =  3240/.  for  the  value  that  was  to  be  found. 

Exavip.  2.  What  annuity,  or  yearly  pension,  being  for 
borne  five  years,  will,  in  that  time,  amount  to,  or  raise  u 
stock  of,  3240/.  at  Af  per  cent,  interest  ? 

In  this  case  we  have  given  n  -=.  5^  r  z=z  .04,  and  m  = 

3240,  and  therefore,  by  theorem  2,  A  (= • 

n'\-^nxn — \Xr 

—  — — -)  =  600  ;  vv^hich  is  the  annuitv  required. 
5  +  4/  '      ^ 

Examp.  3.   At  what  rate  of  interest  will  an  annuity  oi 

560/.  in  seven  years,  raise  a  stock  of  4508/.  ? 

In  th^s  case  we  have  given  A  =  560,  n  =  7,  and  m  = 

1  ,^,        1  r.^  ^  /  2;7^  —  272  A      \ 

4508  ;  whence  (by  theor.  3)  we  have  r  (= ■■■  ) 

nxn —  1  X  A/ 

^  9016  —  7840  ^  ^^^  _  ^^^^  interest  of  1/.  for  one  year  ; 

42  X  560 
therefore  it  will  be,  as  1  :  ,05  :  :  100  :  5  per  cent,  the  rate 
required. 

Examp.  A:.  How  long  must  an  annuity  of  560/.  be  for- 
borne, to  raise  a  r.tock  of  4^08/.  supposing  interest  to  be  ^ 
per  cent.  P 


Of  Interest  and  Annuities^  233 

Here,  we  have  given  A  =  560,  r  =  .05,  m  =  4508 ; 

whence,  by  theorem  4,  we  also  have  /?(=-' )  =19.5 ; 

r  At 

and  consequently  n  (=  y  ^  +  /?2  —  /?)  =  7  ;  which  is 

the  number  of  years  required. 

Note*  If  the  rent  or  pension  be  payable  half-yearly, 
or  quarterly,  the  method  of  proceeding  will  be  still  the 
same,  provided  n  be  always  taken  to  express  the  num- 
ber of  payments,  and  r  the  interest  of  1/.  for  the  time 
in  which  the  first  payment  becomes  due.  Thus,  if  it 
were  required  to  find  what  300/.  half-yearly  pension 
would  amount  to  in  five  years  at  4  per  cent,  interest :  then 
the  simple  interest  of  1/.  for  half  a  year  being  =  ,02,  and 
the  number  of  payments  =  10,  we,  in  this  case,  have  A  = 
300,  r  =  ,02,  and  72  =  10  ;   and  consequently  m  {by  theo-^ 

n  — •  1 
rem  1)  =  r  A  -f-  ?z  x  X  r  A  =  3270/.  which  is  the 

value  sought.     And  the  like  is  to  be  observed  in  what  fol- 
lows hereafter. 


Of  the  Present  Values  of  Annuities,   or  Pensions^ 
computed  at  Simple  Interest. 

"A  =  the  annuity,  pension,  or  yearly  rente 
y  r  =  the  interest  of  lA  for  one  year. 

,      [    /z  =  the  number  of  years. 

^  V  =.  the  present  value  of  the  annuity. 
Then,  because  the  amount  of  the  annuity,  in  n  years, 
is  found  above  to  be  nA.  -^  ^n  .  n  —  1  .  r A,  and  sincfc 
1/.  present  money,  is  equivalent  to  1  +  nr  to  be  re- 
ceived at  the  end  of  the  time  n,  we  therefore  have 
1  +nr  ;  1 ,:  :  wA  +  ^n  .  n  —  1  .  rA  (the  said  amount) 

nK  -\-^n  .  n —  1   .  rA    .  .      ,      , 

:    ^  ,  Its  required  value,  m  present 

money.  But  it  may  be  observed,  that  this  method,  given 
by  authors  for  determining  the  values  of  annuities,  ac- 
cording to  simple  interest^  is,  in  reality,  a  particular  sort, 

2  H 


234  Of  Interest  and  Annuities. 

or  species  of  compound  interest ;  since  the  allowing  of  in- 
terest upon  the  annuity,  as  it  becomes  due,  is  nothing 
less  than  allowing  interest  upon  interest ;  the  annuity  it- 
self being,  properly,  the  simple  interest,  and  the  capital, 
from  whence  it  arises,  the  principal.  It  is  true,  the  sum 
1  -f-  nr^  expressing  the  amount  of  !/•  is  given,  strictly 
speaking,  according  to  simple  interest:  but  the  conclu- 
sion (as  a  late  author  ^  very  justly  observes)  would  be 
more  congruous,  and  answer  better,  were  the  same  al- 
lowances to  be  made  therein  as  are  made  in  finding  the 
amount  of  the  annuity ;  that  is,  were  interest  upon  in- 
terest to  be  taken  once  and  no  more.  Agreeable  to  this 
assumption,  r,  the  interest  of  1/.  being  considered  as  an 
annuity,  its  amount  in  n  years  (by  writing  r  for  A,  in 
the  general  formula  above)  will  be  given  =  nr  -|- 
|?2  .  ;?  •^-  1  .  r" :  to  which  the  principal  1/.  being  add- 
ed, the  aggregate  1  +  nr  +  ^n  .  n  —  1  .  r^  will  there- 
fore be  the  whole  amount  of  1/.  in  the  time  7i;  and  so 
we  shall  have  1  +  nr  -{-  ^n  .  n  —  1  .  r^  :  1  :  :  ?zA  -|- 

^n  •  n  —  1  .  r  A  : — =  ^,  the  true 

2  +  2nr  +  n  ^  7i  —  1  .  r^ 
value  of  the  annuity,  according-  to  the  said  hypothesis. 
From  which  equation  others  may  be  derived,  by  means 
w^hereof  the  different  values  of  A,  w,  and  r,  may  be 
successively  determined.  But,  as  this  method  of  allow- 
ing interest  upon  interest,  once  and  no  more,  is  arbitrary, 
and  the  valuation  of  annuities,  according  to  simple  inte- 
rest, a  matter  of  more  speculation  than  real  use,  it  being 
not  only  customary,  but  also  most  equitable  to  allow  com- 
pound interest  in  these  cases,  I  shall  not  stay  to  exemplify 
it,  but  proceed  to 

The  Resolution  of  the  various  Cases  of  Compound  In- 
teresty  and  of  Annuities^  as  depending  thereon, 

fp  __   f  the  amount  of  1/.  in  one  year,  iyzz.  princi- 
Let    ^      ""  \      pal  and  interest. 

[V  =  any  sum  put  out  at  interest. 

*  Mr.  Hardyy  in  his  Annuities. 


Let  <^ 


OJ  Interest  and  Annuities.  235 

■  n  =  the  number  of  years  it  is  lent  for. 
a  =  its  amount  in  that  time. 
A  =  any  annuity  forborne  n  years. 
m  =z  its  amount. 

___   ( the  present  value  of  the  annuity  for  the  same 
^  ^  ""  I      time. 

Therefore,  since  one  pound,  put  out  at  interest,  in  the 
first  year  is  increased  to  R,  it  will  be,  as  1  to  R,  so  is  R, 
the  sum  forborne  the  second  year,  to  R^,  the  amount  of 
one  pound  in  two  years  ;  and  therefore  as  1  to  R,  so  is 
R^,  the  sum  forborne  the  third  year,  to  R^,  the  amount 
in  three  years :  whence  it  appears  that  R'*,  or  R  raised 
to  the  power  whose  exponent  is  the  number  of  years, 
will  be  the  amount  of  one  pound  in  those  years.  But 
as  1/.  is  to  its  amount  R**,  so  is  P  to  (a)  its  amount,  in 
the  same  time  ;  whence  we  have  P  x  R'*  =  «.  More- 
over, because  the  amount  of  one  pound,  in  n  years,  is 
R»,  its  increase  in  that  time  will  be  R"  — -  1  ;  but  its 
interest  for  one  single  year,  or  the  annuity  answering  to 
that  increase,  is  R  —  1  ;  therefore,  as  R  —  1  to  R'*  —  1, 

A  X  R" 1 

so  is  A  to  m.     Hence  we  get  — rr =  w.     Fur- 

K  —  1 

thermore,  since  it  appears  that  one  pound,  ready  money, 

is  equivalent  to  R'*,  to  be  received  at  the  expiration  of 

A  X  R" 1 

n  years,  we  have,  as  Rn  to  1,  so  is  — (the  sum 

in  arrear)  to  ^,  its  worth  m  ready  money ;  and  therefore 

Axl  — i- 
R« 

R-l       ="• 

From  which  three  original  equations  others  may  be 
derived,  by  help  whereof  the  various  questions  relating  to 
compound  interest,  annuities  in  arrear,  and  the  present 
values  of  annuities^  may  be  resolved. 


235  Of  Interest  and  Annuities. 

Thus,  because  PR'''  is  =  a,  there  will  come  out  P  = 


=p1v 


— ,  and  R  =  -^  I " ,  &c.  or  by  exhibiting  the  same  equations 

in  logarithms  (which  is  the  most  easy  for  practice)  we  shall 
have 

1°.  Log.  a  r=z  log.  P  4-  72  X  log.  R. 

2°.  Log.  P  =  log.  a  —  n  X  log.  R. 

3°.  Log.R=^°S'"-^°g-P. 

?l 

o     ^  _  log.  6?  —  log.  P 

^•^"-      i^R      • 

Which  four  theorems,  or  equations,  serve  for  the  four 
cases  in  compound  interest. 

Again,  since   m  is  =   : HI-.,  we  shall  have 

1°.  Log.  m  =  log.  A  +  log.  R«  —  1  —  log.  R  —  1. 
2°.  Log.  A  =  log.  ?n  —  log.  R" —  1  +  log.  R  —  1. 

o       __  log*  ^^1^  —  m  -f  A  —  log.  A 

^"  ^~—  iS^TR 

4°.  Rn_^  +  ^_1=0. 
A    ^A 

To  which  the  various  questions  relating  to  annuities  in 

arrear  are  referred. 

Moreover,  seeing  A  x  -~ is  =  v^  we  thence  have 

R  —  1 


1 


I"".  Log.  V  =  log.  A  +  log.  1  —  —  —  log.  R  —  1. 

2°.  Log.  A  =  log.  V  +  log.  R  —  1  —  log.  1  —  ^• 
o    ^  _  log.  A  —  log.  A  -f  u  —  vR 

%j   •    Tl  — —  —————— —————^ — • 

log.  R 

4%  R«+^^i:  +  ixR'»  +  ^  =  a 

V  V 


Of  Interest  and  jAnnuities*  237 

The  use  of  which  theorems,  respecting  the  present  va- 
lues of  annuities,  as  well  as  of  the  preceding  ones,  for  com- 
pound interest  and  annuities  in  arrear,  will  fully  appear 
from  the  following  examples. 

Examp.  1.  To  find  the  amount  of  575L  in  seven  years, 
at  four  j&^r  cent,  per  annum^  compound  interest. 
n^  In  this  case  we  have  given  P  =  575^  R  =  1,04,  and 
n-=i7  \  therefore,  hy  theorem  1,  log.~  a  =  log.  S7S  +  7 
log.  1,04  =  2,8789011  ;  and  consequently  a  =  756,66, 
or  7S^L  13^.  ^\d.  the  value  required. 

Examp.  2.  What  principal,  put  to  interest,  will  raise  a 
stock  of  1000/.  in  fifteen  years,  at  5  per  cent.  P 

Here  we  have  given  R  =  1,05,  tz  =  15,  and  a  =  1000  ; 
therefore,  by  theorem  2,  log.  P  =  log.  1000  —  15  log.  1,05 
=  2,6821605  ;  and  consequently  P  =  481,02  or  481/,  0^. 
4|c/.  the  value  sought. 

Examp.  3.  In  how  long  time  will  57 SL  raise  a  stock  of 
756/.  13^.  2\d.  at  4<  per  cent.P 

In  this  case  we  have  R  =  1,04,  P  =  575^  and  a:= 
756,66;  whence,  bytheor.4,  ^^^og- 756,^6 -log- 575 

log.  1,04 
=  7,  the  number  of  years  required. 

Examp.  4.  To  find  at  what  rate  of  interest  481/.  in  fif- 
teen years,  will  raise  a  stock  of  1000/. 

Here  we  have  given  P  =  481,  a  =  1000,  and  w  =  15  ; 

therefore,  by  theorem  3,  log.  R  =  —^^ ^^' 

=  .  0211903,  whence  R  =  1,05  ;  consequently  5  per  cent. 
is  the  rate  required. 

The  four  last  examples  relate  to  the  cases  in  compound 
interest ;  the  four  next  are  upon  the  forbearance  of  an- 
nuities. 

Examp.  1.  If  50/.  yearly  rent,  or  annuity,  be  forborne 
seven  years,  what  will  it  amount  to,  at  4  per  cent,  per  an- 
num^ compound  interest  ? 

Here  we  have  R  =  1,04,  A  =  50,  and  tz  =  7  5    and 


238  Of  Interest  and  Annuities* 


therefore,  by  theor,  1 ,  log.  m  (=  log.  A  +  log.  R"  —  1 

—  log.  R—  1)  =  log.  50  +  log.  Ifiiy  —  1,  —  log.  ,04 
=  2,596597 ;  and  consequently  7n  =  395/.  the  value  that 
was  to  be  found. 

Examp.  2.  What  annuity,  forborne  seven  years,  will 
amount  to,  or  raise  a  stock  of  395/.  at  4  per  cent,  com- 
pound interest  ? 

In  this  case  we  have  given  R  =  1,04,  n  =  7,  and 
m  =  395  ;    whence,   hij  theorem  2,1  og.   A   (  =  log.  m 

—  log.  R*«— 1  +  log.  R  — 1)  =  log,  395  —  log.  1^] 7  —  1 
+  log.  ,04  =  1,6989700;  and  consequently  A  =  50/.  which 
is  the  annuity  required. 

Examp.  3.  In  how  long  time  will  50/.  annuity  raise  a 
stock  of  395/.  at  4  per  cent,  per  annum^  compound  inte- 
rest? 

Here  we  have  R  =  1,04,  A  =  50,  w  =  395  ;    and 

therefore,  hii  theor.  3,  n  (=  -Sli^! —      l^^  "^^  ^^'      ) 

^  ^  log.  R  ^ 

:=::  1_ =  7  the  number  of  vears  required. 

,0170333 

Exam  A  4.  If  120/.  annuity,  forborne  eight  years, 
amount  to,  or  raise  a  stock  of  1200/.  what  is  the  rate 
of  interest  ? 

In  this  case  we  have  given  n  =  8,  A  =  120,  and  m 
■=.  1200,  to  find  R ;  therefore,  by  theorem  4,  we  have 
R»  —  loR  +  9  =  0,  from  which,  by  any  of  the  methods 
in  sect.  13,  the  required  value  of  R  will  be  found  = 
1,06287;  therefore  the  rate  is  6,287,  or  6/.  5s.  9d.  per 
cent,  per  annum. 

The  solution  of  the  last  case,  where  the  rate  is  re- 
quired, being  a  little  troublesome,  I  shall  here  put  down 
an  approximation  (derived  from  the  third  general  for- 
mula^ at  p.  165)  which  will  be  found  to  answer  very 
near  the  truth,  provided  the  number  of  years  is  not  very 
great. 


Of  Interest  and  Annuities »  239 


Let  Q  = ;  then  will 

2  .  w^  —  nA. 

3000Q  +  2n—l.  400 

be  the  rate 


6Q.5Q  +  37Z  — 4  +  I-.  n  —  2  .  tin— 13 
per  cent,  required. 


Thus,  for  example,  let  n  =  8,  A  =  120,  and  m  =  1200  ; 
m  will  Q  = 
42000  +  6000 


then  will  Q  =  '- =14,  and  the  rate  itself 

2   .  240  ' 


84  X  90  +  75 


=  6,287,  as  above. 


The  preceding  examples  explain  the  different  cases  6i 
annuities  in  arrear  ;  in  the  following  ones  the  rules  for  the 
valuation  of  annuities  are  illustrated. 

Examp.  1.  To  find  the  present  value  of  100/.  annuity, 
to  continue  seven  years,  allowing  4  per  cent,  per  annmri^ 
compound  interest. 

Here  we  have  given  R  =  1,04,  A  =  100,  and  n  =  7 ; 
and  therefore,    by   theorem    1,    log.    t;    ( =    log.    A   + 

log.    1   —  ~    —  log.   R  —   1  )    =    log.    100    +   log. 

1  — log,  ,04  =  2,778296  j  and  consequently 

1,04^ 
t;  =  600,2  =  600/.  4sy.  which  is  the  value  that  was  to  be 
found. 

Examp.  2.  What  annuity,  or  yearly  income,  to  con- 
tinue 20  years,  may  be  purchased  for  1000/.  at  3^  per 
cent.  P 

In  this  case,  R  =  1,035,  ;z  =  20,  v  =  1000; 
whence,    by    theorem    2,    we    have    log.   A   ( =  log.  v 


+  log.  R  —  1  _  log.  1  _      )  =  1,847336;    and  con- 
sequently  A  =  70,36,  or  70/.  7s.  %d. 


240  Of  Interest  and  Annuhies. 

Examp.  3.  For  how  long  time  may  one,  with  600/.  pur- 
chase an  annuity  of  100/.  at  4/?^r  cent.  ? 

In  this  example,  we  have  R  =  1,04,  A  =  100,  and 
V    =    600 ;      and     therefore,     by    theorem     3,     n     (  = 

log.  A  — log.  A  4-  ^  —  ^^Rn        ^   ,,  u        c 

—2 ^ — .^ A  ~  7   the  number  of  years 

log.  K 
required* 

Examp.  4.  To  determine  at  what  rate  of  interest  an 
annuity  of  50/.  to  continue  ten  years,  may  be  purchased, 
for  400/. 

Here  A  =  50,  n  =  10,  and  v  =  400;    whence,  by 

A  A 

theorem  4,  R'^+i (-lxR"H being  =  0,  we 

have  R"  —  1,125R^°  +  ,125  =  0;  which  equation 
resolved,  gives  the  required  value  of  R  =  1,042775  ; 
and  consequently  the  rate  of  interest  4,2775/.  per  an-^ 
num. 

The  solution  of  this  last  case  being  somewhat  tedious, 
the  following  approximation  (which  will  be  found  to  an- 
swer very  near  the  truth,  when  the  number  of  years  is  not 
very  large)  may  be  of  use. 


Assume  Q  =  — 1— Jt — '- —  :  so  shall 

2nK  —  2v 

3000Q  —  2/1 -f-  1  X  400 
express  the 


6Q.  5Q— 3?2  — 4  +  1  .  n  +  2  .   ll/z  +  lS 
rate  per  cent,  very  nearly. 

Thus,  for  example,  let  A  (as  above)  be  =  50,  n  =  10, 

andi;  =  400;  then,  Q being  =i^^  ^^  ^/^     =    27.5, 
'  ^  1000  —  800 

,  82500  —  8400  .  ^^^^    r      ^-u         ^    ^.^ 

we  have ,  or  4,2775,  for  the  rate  per 

165  X  103,5  +  246'  '  ,^ 

'  cent,  the  same  as  before. 


Of  Plane  Trigonometry. 


241 


SECTION  XVII. 


Of  Plane  Trigonometry* 

DEFINITIONS. 

1.  PLANE  trigonometry  is  the  art  whereby,  hav- 
ing given  any  three  parts  of  a  plane  triangle  (except 
the  three  angles),  the  rest  are  determined.  In  order  to 
which,  it  is  not  only  requisite  that  the  peripheries  of  cir- 
cles, but  also  that  certain  right  lines,  in  and  about  the 
circle,  be  supposed  divided  into  some  assigned  number  of 
equal  parts. 

2.  The  periphery  of  every  circle  is  supposed  to  be 
divided  into  360  equal  parts,  called  degrees  ;  and  each 
degree  into  60  equal  parts,  called  minutes  ;  and  each 
minutes  into  60  equal  parts,  called  seconds,  or  second- 
minutes,  £sfc.  Any  part  of  the  periphery  is  called  an 
arch,  and  is  measured  by  the  number  of  degrees  and 
minutes,  c£?c.  it  contains. 

3.  The  difference  of  any  arch  from  90  degi'ees,  or  a  , 
quadrant,  is  called  its  complement,  and  its  difference  from 
180  degrees,  or  a  semicircle,  its  supplement. 

4.  A  chord,  or  sub- 
tense, is  a  right  line 
drawn  from  one  ex- 
tremity of  an  arch  to 
the  other ;  thus  BE 
is  the  chord  or  sub- 
tense of  the  arch 
BAE  or  BDE. 

5.  The  sine  (or 
right-sine)  of  an  arch 
is  a  right  line  drawn 
from  one  extremity 
of  the  arch  perpen- 
dicular to  the  diame- 
ter passing  through  the  other  extremity:  thus,  BF  is  the 
sine  of  the  arch  AB,  or  BD. 

21 


242  Of  Plane  Trigonometry. 

6.  The  versed- sine  of  an  arch  is  the  part  of  the  diame- 
ter intercepted  between  the  arch  and  its  sine :  so  AF  is 
the  versed-sine  of  AB,  and  DF  of  DB. 

7.  The  co-sine  of  an  arch  is  the  part  of  the  diameter 
intercepted  between  the  center  and  the  sine  ;  and  is  equal 
to  the  sine  of  the  complement  of  that  arch.  Thus,  CF  is 
the  co-sine  of  the  arch  AB,  and  is  equal  to  BI,  the  sine  of 
its  complement  HB. 

8.  The  tangent  of  an  arch  is  a  right  line  touching 
the  circle  in  one  extremity  of  that  arch,  continued  from 
thence,  to  meet  a  line  drawn  from  the  center  through  the 
other  extremity  ;  which  line  is  called  the  secant  of  the 
same  arch  :  thus  AG  is  the  tangent,  and  CG  the  secant  of 
the  arch  AB. 

9.  The  co-tangent  and  co-secant  of  an  arch  are  the 
tangent  and  secant  of  the  complement  of  that  arch  :  thus 
HK  and  CK  are  the  co-tangent  and  co-secant  of  the  arch 
AB. 

10.  A  trigonometrical  canon  is  a  table  exhibiting  the 
lengths  of  the  sine,  tangent,  csfc.  to  every  degree  and  mi- 
nute of  the  quadrant,  with  respect  to  the  radius,  which  is 
supposed  unity,  and  conceived  to  be  divided  into  10000000 
or  more  decimal  parts.  Upon  this  table  the  numerical 
solution  of  the  several  cases  in  trigonometry  depends  ;  it 
will  therefore  be  proper  to  begin  with  its  construction. 


PROPOSITION  I. 

The  number  of  degrees  and  minutes^  &c.  in  an  arch  be-^ 
ing  given  ;  to  find  both  its  sine  and  co-sine. 

This  problem  is  resolved,  by  having  the  ratio  of  the 
circumference  to  the  diameter,  and  by  means  of  the 
known  series  for  the  sine  and  co-sine  (hereafter  de- 
monstrated). For,  the  semi-circumference  of  the  circle, 
whose  radius  is  unity,  being  3,141592653589793,  £5?r. 
it  will  therefore  be,  as  the  number  of  degrees  or  mi- 
nutes in  the  whole  semicircle  is  to  the  degrees  or 
minutes  in  the  arch  proposed,  so  is  3,14159265358,  ^c. 
to  the  length  of  the  said  arch ;  which  let  be  denoted  by 
a ;  then,  by  the  series  above  quoted^  its  sine  will  be  ex- 


Of  Plane  Trigonometry*  245 

pxessed  by  a  -  ^  +  ^-^^  -  ^  .  3  . /'o  .  6  .  7' 
&fc*  and  its  co-sine  by  1  — 


2  ^4  ^6 


2       2.3,4       2.3.4.5.6 

^  2 .  3  .  4  .  5  .  6  .  r  .  8 

Thus,  for  example,  let  it  be  required  to  find  the  sine 
of  one  minute:  then,  as  10800  (the  minutes  in  180  de- 
grees) :  1  :  :  3,14159265358,  i^c.  :  .000290888208665 
=  the  length  of  an  arch  of  one  minute :     therefore,  in 

this    case,    a  =  .000290888208665,    and    -^    ( =  ^) 

^  •  o  O 

=  .000000000004102,  &fc.     And,  consequently, 
.000290888204563  =  the  required  sine  of  one  minute* 

Again,  let  it  be  required  to  find  the  sine  and  co-sine 
of  five  degrees,  each  true  to  seven  places  of  decimals. 
Here  ,0002908882,  the  length  of  an  arch  of  1  minute 
(fpund  above),  being  multiplied  by  300,  the  number  of 
minutes  in  5  degrees,  the  product  .08726646  will  be  the 
length  of  an  arch  of  5  degrees  :  therefore,  in  this  case  we 
have 

a  =       ,08726646, 

—|.=  —, 0001 1076, 
o 

a^ 

+ =  +  ,00000004, 

^120  ' 

i^c>  and  consequently  ,08715574  =  the  sine  of  5  degrees. 

Also, 

—  =,00380771, 
2 

a* 

—  =  ,00000241 ; 
24      ' 

^and  consequently  ,9961947  =  th^  co-sine  of  5  degrees. 
After  the  same  manner,  the  sine  and  co-sine  of  any 
other  arch  may  be  derived  ;  but  the  greater  the  arch  is, 
the  slower  the  series  will  converge,  and  therefore  a  greater 
number  of  terms  must  be  taken  to  bring  out  the  conclusion 
to  the  same  degree  qi  exactness. 


244 


Of  Plane  Trigonometry. 


But  there  is  another  method  of  constructing  the  trigo- 
nometrical canon  ;  which,  though  less  direct,  is  more  geo- 
metrical ;  and  that  is  by  determini^ng  the  sines  and  tan- 
gents of  different  arches,  one  from  another,  as  in  the  en- 
suing propositions. 


PROPOSITION  IL 

The  sine  of  an  arch  being  given;  to  find  its  co-sine^  tan- 
gent^ co-tangent^  secant ^  and  co-secant. 

Let  AE  be  the  proposed   arch,  EF  its  sine,  CF  its 


co-sme, 


AT  its  tangent,  DH  its  co-tangent,  CT  its 
secant,  and  CH  its  co-secant :  then  {by  Euc.  47.  1.)  we 
shall  have    CF  =  VCE^  —  EF^ ;      from    whence    the 

co-sine  will  be  known ;  and 
then,  by  reason  of  the  similar 
triangles,  CFE,  CAT,  and 
CDH,  it  will  be, 

1.  CF  :  FE  :  :  CA  :  AT; 
whence  the  tangent  is  known. 

2.  CF  :  CE   :   :  CA  :  CT ; 
whence  the  secant  is  known. 

3.  EF  :  CF  :  :  CD  :  DH  ; 
whence  the  co-tangent  is  known. 


CD  :  CH  ;  whence  the  co-secant  is  also 


c 

4.  EF 
known. 

Hence  it  appears, 

1.  That  the  tangent  is  a  fourth  proportional  to  the  co- 
sine, the  sine,  and  the  radius. 

2.  That  the  secant  is  a  third  proportional  to  the  co-sine 
and  the  radius. 

3.  That  the  co-tangent  is  a  fourth  proportional  to  the 
sine,  the  co-sine,  and  the  radius. 

4.  That  the  co-secant  is  a  third  proportional  to  the  sine 
and  the  radius. 

5.  And  that  the  rectangle  of  the  tangent  and  co-tangent 
is  equal  to  the  square  of  the  radius. 


Of  Plane  Trigonometry*  245 

PROPOSITION  III. 

The  co-sine  CF  of  an  arch  AE  being  given  ;  to  find  the 

ine  and  co-sine  of  half  that  arch* 

From  the  two  extremities  of  the  diameter  AB  draw 

the  subtenses  AE  and  BE  ;  and  let  CQ  bisect  the  arch 

AE  in  Q  and  its  chord  (perpendicularly)  in  D  ;    then, 

since  the  angle  BE  A  is  a  right  one-  (%  Euc.  31.  3.),  the 

triangles     ABE     and  ^^^^ ^^JE 

ADC  are  similar  ; 
and,  therefore,  AC 
being  =  ^  AB,  AD 
must  be  =  i  AE,  and 
CD  =  \  BE  :  but  AE 
is  =  VAB  X  AF,  and 
BE  =  VAB  X   BF  ;  therefore 


AD  =  iVAB_  __ 

CD  =  Iv^AB  xW  =  Vp^  X~BF  =  the  co-sine 

Hence  it  is  evident,  that  the  sine  of  the  half  of  any 
arch  is  a  mean  proportional  between  the  half  radius  and 
the  versed-sine  of  the  whole  arch  ;  and  its  co-sine  a, mean 
proportional  between  half  the  radius  and  the  versed-sine  of 
the  supplement  of  the  same  arch. 

PROPOSITION  IV. 

The  sine  AD,  and  co-sine  CD,  of  an  arch  AQ  being 
given  ;  to  find  EF,  the  sine  of  the  double  of  that  arch.  {See 
the  preceding  fgure)* 

Since  AE  =  2AD,  and  BE  =  2CD,  and  the  triangles 
ABE  and  AEF  are  alike  {bif  Euc*  8.  6.),  we  have,  as 
AB  (2AC)  :  AE  (2AD)  :  :  BE  (2CD)  :  EF  ;  whence 
it  appears,  that  the  sine  of  double  any  arch  is  a  fourth 
proportional  to  the  radius,  the  sine,  and  the  double-co- 
sine of  the  same  arch. 

PROPOSITION  V. 

The  sine  CD,  and  tangent  BE,  of  a  very  small  arch^are 
nearly  in  the  ratio  of  equality* 

For,   the  triangles  ADC    and   ABE   being   similar. 


246  OfPlmc  Trigonometry. 

thence  will  AD  :  AB  :  :  DC  :  BE  :  but  as  the  point  C 
approaches  to  B,  the  difference  of  AB  and  AD  will 
become  indefinitely  small  in  respect  of  AB,  and  there- 
fore the  difference  of 
BE  and  DC  will  liker 
wise  become  indefinitely 
small  with  respect  to  BE 
or  DC. 

Corollary.  Because  any 
arch  BC  is  greater  than 
its  sine  and  less  than  its 
tangent  ;  and  since  the 
sine  and  tangent  of  a  very  small  arch  are  proved  to  be 
nearly  equal,  it  is  manifest  that  a  very  small  arch  and  its 
sine  are  also  nearly  in  the  ratio  of  equality. 

PROPOSITION  yi. 

To  find  the  sine  of  an  arch  of  one  minute* 

The  sine  of  30  degrees  is  known,  being  half  the 
chord  of  60  degrees,  or  half  the  radius  ;  therefore,  by  prop. 
2.  and  3.  the  sine  of  15  degrees  will  be  known  ;  and,  the 
sine  of  15  degrees  being  known,  the  sine  of  7""  30'  will 
be  found  {by  the  same  propositions)^  and  from  thence  the 
sine  of  3°  45' ;  and  so  likewise  the  sine  of  half  this  ;  and 
^o  on,  till  12  bisections  being  made,  we  come,  at  last, 
to  .the  sine  of  an  arch  of  52",  44'^^  03';",  45'"";  which 
sine  (by  corol.  to  the  preceding  prop.)  will,  as  the  co-sine 
is  nearly  equal  to  the  radius,  be  nearly  equal  to  the  arch 
itself.  Therefore  we  have,  as  52",  44"',  03"",  45""',  is 
to  1',  so  is  the  length  of  the  former  of  these  arches  (found 
as  above)  to  the  length  of  an  arch  of  one  minute,  or  t!iat 
of  its  sine,  very  nearly. 

If  it  be  taken  for  granted,  that  3,1415926535,  ^c. 
is  the  length  of  half  the  periphery  of  the  circle  whose 
radius  is  unity,  we  shall  have,  as  10800,  the  number  of 
minutes  in  180°,  or  the  whole  semicircle,  is  to  one  mi- 
nute, so  is  3,1415926535,  &fc.  the  whole  semicircle,  to 
0,000290888208,  the  length  of  an  arch  of  one  mijnute,  or 
that  of  its  sine,  very  nearly.  1^ 


Of  Plane  Trigonometry, 


24>7 


PROPOSITION  VIL 

If  there  he  three  equidifferent  arches  AB,  BC,  and  AD, 

it  will  be^  as  the  radius  is  to  the  co-sine  of  their  common 
difference  BC  or  CD,  so  is  the  sine  CF,  of  the  mean^  to 
half  the  sum  of  the  sines  BE  +  DG,  of  the  two  extremes  ; 
and  as  the  radius  is  to  the  sine  of  the  common  difference^  so 
is  the  co-sine  FO  of  the  mean^  to  half  the  difference  of  the 
sines  of  the  two  extremes. 

For,  let  BD  be  drawn,  cutting  the  radius  OC  in  7n ; 
also  draw  mn  parallel  to  CF,  meeting  AO  in  7Z,  and 
BH  and  77iv  parallel  to  AO,  meeting  DG  in  H  and  v : 
then,  because  the  arches  BC  and  CD  are  equal  tQ  each 


A  E 


other,  OC  is  not  only  perpendicular   to    BD,  but  also 
bisects  it  (^Euc.  3.  3.)  ;  whence  it  is  evident  that  Bm,  or 
Dw,  will  be  the  sine  of  BC  or  CD,  and  Om  its  co- 
sine ;  and  that  mn^  being  an  arithmetical  mean  between 
the  sines   BE   and  DG,  of  the  two  extremes,  is  equal 
to  half  their  sum,  and  T^v  equal  to  half  their  difference. 
Moreover,  by  reason  of  the  similarity  of  the  triangles 
OCF,  OwiTz,  and  Dmv,  it  will 
be  as  OC  :  Om  \  \  C¥  ,  mn     \  q,    ^    j^ 
and  as  OC  :  Dw  :  :  FO  :  D^  /  ^'  ^-  ^' 
COROL.  1. 
Since,    from    the    foregoing    proportions,    mn    is    =: 

Om  X  CF  Dm  X  FO    .    .        . ,    ^ 

- ,  and  U'o  (=  z;H)  =  — ~^^ ,  it  is  evident 


OC 


OC 


248  Of  Plane  Trigonometry. 

that  DG  (=  mil  +  D.)  will  be  =  Q>»xCF  +  DmxFO 

and  BE  (=  mn  —  va)  = -— :  from 

whence  it  appears  that  the  sine  (DG)  of  the  sum  (AD) 
of  any  two  arches  (AC  and  CD)  is  equal  to  the  sum  of 
the  rectangles  of  the  sine  or  the  one  into  the  co-sine 
of  the  other,  alternately,  divided  by  the  radius ;  and 
that  the  sine  (BE)  of  their  difference  (AB)  is  equal  to 
the  difference  of  the  same  rectangles,  divided  also  by  the 
radius, 

COROL.  2. 

20w  X  OF 


Moreover,  seeing  DG  +  BE  (2w?2)  is  = 


OC 


and  DG  —  BE  (=  DH  =  2D^)  =  !H^-jJ!2,  from  the 
former  of  these,  we  have  DG  =  — ^^-~ BE,  and 

from  the  latter,  DG  = — ^^ +  BE  ;    which  equa- 

tions,  expressed  in  words,  give  the  following  theorems, 

Theor.  1.  If  the  sine  of  the  mean  of  three  equidtfferent 
arches  (supposing  the  radius  unity)  be  multiplied  by  twice 
the  cosine  of  the  common  difference^  and  the  sine  of  either 
extreme  be  subtracted  from  the  product^  the  remainder  will 
be  the  sine  of  the  other  extreme. 

Theor.  2.  Or,  if  the  co-sine  of  the  mean  be  multiplied  by 
ttvice  the  sine  of  the  common  differ ence^  and  the  product  be 
added  to  or  subtracted  from  the  sine  of  one  of  the  extremes^ 
the  sum  or  remainder  xvill  be  the  sine  of  the  other  extreme. 
These  two  theorems  are  of  excellent  use  in  the  con* 
struction  of  the  trigonometrical  canon;  for,  supposing 
the  sine  and  co-sine  of  an  arch  of  1  minute  to  be  found, 
by  prop.  6  and  1 ,  and  to  be  denoted  by  p  and  y,  respec- 
tively ;  then,  the  sine  of  2  minutes  being  given  from 
prop.  4,  the  sine  of  3  minutes  will  from  hence  be  known, 
being  ==  2^  X  sine  2'  —  sine  1'  (by  theor.  l)  or  =  2p 
X  co-sine  of  2'  +  sine  of  I'  (by  theor.  2).     After  the  same 


Of  Plane  Trigonometry » 


249 


inanner  the  sine  of  4'  will  be  found,  being  =  2§^  X  sine  of 
3'  —  sine  of  2',  or  =  2/?  X  co-sine  of  3'  +  sine  of  2'.  And 
thus  the  sines  of  5,  6,  7,  &fc.  minutes  may  be  successively 
derived  by  either  of  the  theorems  ;  but  the  former  is  the 
most  commodious. 

If  the  mean  arch  be  45°,  then,  its  co-sine  "being  = 
V|^,  it  follows  (yr(?m  theorem  2)  that  the  sine  of  the  ex- 
cess of  any  arch  above  45°,  multiplied  by  2V|.  or  \/2, 
gives  the  excess  of  the  sine  of  this  arch  above  that  of 
another  arch  as  much  below  45° ;  thus,  VS  x  sine  of 
10°  =  sine  of  SS""  ■—  sine  of  ZS"" ;  and  Vi  x  sine  of  15° 
=  sine  of  60°  —  sine  of  30° ;  and  so  of  others  :  which  is 
useful  in  finding  the  sines  of  arches  greater  than  45°. 

But,  if  the  mean  arch  be  60  degrees,  then  its  co-sine 
being  i,  it  is  evident,  from  the  same  theorem^  that  the 
sine  of  the  excess  of  any  arch  above  60°,  added  to  the 
sine  of  another  arch  as  much  below  60°,  will  give  the 
sine  of  the  first  arch,  or  greater  extreme :  thus,  the  sine 
of  10°  +  sine  50°  =  sine  70°,  and  sine  15°  -f-  sine  45°  = 
sine  IS"^ ;  from  whence  the  sines  of  all  arches  above  60 
degrees,  those  of  the  inferior  arches  being  known,  are  had 
by  addition  only* 

PROPOSITION  VIII. 


In  dny  right-angled  plane  triangle  ABC,  it  will  fe,  as 
the  base  AB  is  to  the  perpendicular  BC,  50  is  the  radius  {of 
the  tables)  to  the  tangent  of  the  angle  at  the  base. 

Let  DA  be  the  radius  to  which  the  table  of  sinea 
and  tangents  is  a- 
dapted,  and  DE  the 
tangent  of  the  an- 
gle A  ;  then,  by  rea- 
son of  the  similarity 
of  the  triangles  ABC 
and  ADE,  it  will  be, 
as  AB  :  BC  :  :  AD 
:  DE.     %  E.  D. 


250 


Of  Plane  Trigonometry, 


PROPOSITION  IX. 

In  every  plane  triangle^  it  will  he^  as  any  one  side  is  to 
the  sme  of  the  opposite  angle^  so  is  any  other  side  to  the  sine 
of  its  opposite  angle. 

For,  let  ABC  be  the  proposed  triangle;  take  CF  = 
AB,  and  upon  AC  let  fall  the  perpendiculars  BD  and 
EF  ;    which  will  be  the  sines  of  the  angles  A  and  C, 

to  the  equal  ra- 
dii AB  and  CF. 
But  the  trian- 
gles CBD  and 
CFE  are  simi- 
lar, and  there- 
fore CB  :  BD  :  : 
CF  (AB)  :  FE  ; 
that  is,  as  CB  is  to  the  sine  of  A,  so  is  AB  to  the  sine  of 
C.     S>.£.D. 


PROPOSITION  X. 

In  every  pla7ie  triangle^  it  will  be^  as  the  sum  of  any  two 
sides  is  to  their  difference^  so  is  the  tangent  of  the  complex 
ment  of  half  the  angle  included  by  those  sideSj  to  the  tan- 
gent  of  the  difference  of  either  of  the  other  two  angles  and 
the  said  complement. 

For,  let  ABC  be  the  triangle,  and  AB  and  AC  the 
two  proposed  sides ;  and  upon  A,  as  a  center,  with  the 
radius  AB,  let  a  semicircle  be  described,  cutting  CA 
produced,  in  D  and  F  ;  so  that  CF  may  express  the  sum, 

and  CD  the 
difference  of 
the  sides  AC 
and  AB  ;  join 
F,  B,  and  B, 

D,  and  draw 

P  AD  ^       DE      parallel 

to  FB,  meeting  BC  in  E  ;  then  the  angle  FBD  being 
aright  one  {by  Euc.  31.  3.)  ADB  will  be  the  comple- 
ment of  the  angle  F,  which  is  equal  to  half  the  pro- 
posed angle  A  (by  Euc.  20.  3.).     Moreover,  seeing  the 


Of  Plane  Trigonormtry. 


251 


angles  FBD  and  EDB  are  both  right  ones,  for  EDB 
is  =  FBD  (=  a  right  angle),  because  DE  is  parallel  to 
FB,  it  is  plain,  that,  if  BD  be  made  the  radius,  BF 
will  be  the  tangent  of  BDF,  and  DE  the  tangent  of 
DBE  :  but,  because  of  the  similar  ti*iangies  CFB  and 
CDE,  CF  :  CD  :  :  BF  :  DE  ;  that  is,  as  the  sum  of  the 
sides  AC  and  AB  is  to  their  difference,  so  is  the  tan- 
gent of  BDF  to  the  tangent  of  DBC  \  which  angle  is  ma- 
nifestly the  excess  of  ABC  above  BDF,  or  ABD  ;  and 
also  the  excess  of  ADB  above  ACB.     ^E.  D. 


PROPOSITION  XL 

As  the  base  of  any  plane  triangle  is  to  the  sum  of  the  two 
sides^  so  is  the  difference  of  the  sides  to  the  difference  of  the 
segments  of  the  base^  made  by  a  perpendicular  falling  from 
the  vertical  angle •  < 

For,  let  ABC  be  the  proposed  triangle,  and  BD  the 
perpendicular ;  from  B  as  a  center,  with  the  interval 
BC,  let  the  circumference  of  a  circle  be  described,  cut- 
ting the  base  AC 
in  G,  and  the  side 
AB,  produced,  in 
F  and  E  :  then 
wall  AE  be  the 
sum  of  the  sides, 
AF  their  differ- 
ence, and  AG  the 
difference  of  the 
segments  of  the 
base  AD  and  DC: 
but  (^by  Eiic.   36. 

3.)  AE  X  AF  =  AC  X  AG  ;  and  therefore  AC  :  AE 
AP:  AG.     %E.D. 


Z52  Of  Plane  Trigonometry, 

The  solution  of  the  cases  of  right-angled  plane  trimigks. 


Given. 

Sought. 

Proportion. 

1 
2 

Thehypothe- 
nuse  AC  and 
the  angles. 

The  leg 
BC. 

As  the  !  adius  (or  the  sine  of  B)  is 
to  the  hyp.  AC ;  so  is  the  sine  of 
A,  to  its  opposite  side  BC.  (^by 
firo/i.  9  ) 

The    hypoth. 
AC  and   one 
leg  AB. 

The 
angles. 

As  AC  :  rad.  :  :  AB  :  sine  of  C ; 
whose  complement  gives  the  an- 
gle A. 

3 

The    hypoth. 
AC  and   one 
le^  AB. 

Theother 
leg  BC. 

i^et  the  angles  be  found  by  case  2  ; 
then,  as  rad.  :  AC  :  :  sme  of  A  : 
BC  {by  firo/i.  9,) 

4 
5 
6 

The      angles 
and    one  leg 
AB. 

The  hy- 

pothenuse 

AC. 

As  sine  of  C  :  AB  :  :  rad.  (sine  of 
B):  AC  (iby  prop.  9, ) 

The      angles 
and   one    leg 
AB. 

The  other 
leg  BC. 

As  sine  of  C  :  AB  :  :  sine  of  A  : 

.     BC  {by  prop.  9.) 

Or,  rad.  :  tang,  of  A  :  :  AB  :  BC 

(byfirofi   8.) 

The  two  legs 
AB  and  BC. 

The 
angles. 

As  AB  :  BC  :  :  rad.  :  tang,  of  A 
{by  prop,  8) ;  whose  complement 
gives  the  angle  C. 

7 

The  two  legs 
AB  and  BC. 

The     hy- 

pothenuse 

AC. 

Find  the  angles  by  case  6,  and 
from  thence  the  hyp.  AC,  by 
case  4. 

Of  Plane  Trigonometry.  253 

The  solution  of  the  cases  of  oblique  plane  triangles. 


Given. 

Sought. 

Proportion. 

1 

rhe    angles 
and  one  side 
AB 

Either  of  the 
other   sides, 
suppose  BC. 

As  sine  of  C  :  AB  :  :  sine  of  A  : 
BC  (dy  firo/i.  9.) 

2 

Two  sides 
AB,  BC  and 
the  angle  C 
opposite  to 
one  of  them. 

The  other 
angles     A 
and  ABC 

As  AB  :  sine  of  C  :  :  BC  :  sine  of 
A ;  which  added  to  C,  and  the 
sum  subtracted  from  180°,  gives 
the  angle  ABC. 

3 

Tsvo     sides 
AB,  BC  and 
the  angle  C 
opposite    to 
one  of  them 

The  other 
side  AC. 

Find  the  angle  ABC  by  case  2  ; 
then,  as  sine  of  A  :  BC  :  :  sine 
of  ABC  :  AC. 

4 
5 

Two   sides 
AB,  AC  and 
the  included 
angle  A. 

The  other 
angle    C 
and  ABC. 

As  AB+ AC  :  AB— AC  :  :  tang,  of 
the  comp.ofl  A  :  tang,  of  an  ang. 
which  added  to  the  said  com.  gives 
the  greater  ang.  C ;  and  subtracted 
leaves  the  lesser  ABC  (firo/i,  10.) 

Two     sides 
AB,  AC  and 
the  included 
angle  A. 

The  other 
side  BC. 

Find  the  angles  by  case  4;  and 
then  BC,  by  case  1 . 

6 

All  the 
sides. 

An  angle, 
suppose  A 

Let  fall  a  perpendicular  BD,  opposite 
the  required  angle,  and  suppose  DG 
=  AD  ;  then  (by prop.  11.)  AC  :  BC 
+  B A  :  :  BC  —  BA  :  CG,  which 
subtracted  from  AC,  and  the  re- 
mainder divided  by  2,  gives  AD  ; 
whence  A  will  be  found,  by  case  2, 
of  right  angles. 

254  The  Application  of-Aigehr 


SECTION  XVIII. 


The  Application  of  Algebra  to  the  Solution  of  Geo- 
metrical Problems. 

WHEN  a  geometrical  problem  is  proposed  to  bfe  re- 
solvecUDy  algebra,  you  are,  in  the  first  place,  to  describe 
a  figure  that  shall  represent  or  exhibit  the  several  parts  or 
conditions  thereof,  and  look  upon  that  figure  as  the  true 
one  J  then,  having  considered  attentively  the  nature  of 
the  problem,  you  are  next  to  prepare  the  figure  for  a  so- 
lution (if  need  be),  by  producing  and  drawing  such  lines 
therein  as  appear  most  conducive  to  that  end.  This  done, 
let  the  unknown  line,  or  lines,  which  you  think  will  be  the 
easiest  found  (whether  required  or  not),  together  with 
the  known  ones  (or  as  many  of  them  as  are  requisite),  be 
denoted  by  proper  symbols  ;  then  proceed  to  the  opera- 
tion, by  observing  the  relation  that  the  several  parts  of 
the  figure  have  to  each  other ;  in  order  to  which,  a  com- 
petent knowledge  in  the  elements  of  geometry  is  abso- 
lutely necessary. 

As  no  general  rule  can  be  given  for  the  drawing  of  lines*, 
and  electing  the  most  proper  quantities  to  substitute  for, 
so  as  always  to  bring  out  the  most  simple  conclusions  (be- 
cause different  problems  require  different  methods  of  so- 
lution), the  best  way,  therefore,  to  gain  experience  in  this 
matter  is  to  attempt  the  solution  of  the  same  problem  se- 
veral ways,  and  then  apply  that  which  succeeds  best  to 
other  cases  of  the  same  kind,  when  they  afterwards  occur. 
I  shall,  however,  subjoin  a  few  general  directions,  w^hich 
will  be  found  of  use. 

1°.  In  preparing  the  figure,  by  drawing  lines,  let  them 
be  either  parallel  or  perpendicular  to  other  lines  in  the 
figure,  or  so  as  to  form  similar  triangles  ;  and  if  an  angle 
be  given,  let  the  perpendicular  be  opposite  to  that  angle-, 
and  also  fall  from  the  end  of  a  given  line,  if  possible. 

2°.  In  electing  proper  quantities  to  substitute  for,  let 
thoge  be  chosen    (whether    required  or  not)    which  lie 


p 


ro  Geometrical  Problems.  25S 

neatest  the  known  or  given  parts  of  the  figure,  and 
by  help  whereof  the  next  adjacent  parts  may  be  expressed, 
without  the  intervention  of' j  surds,  by  addition  and  sub- 
traction only.  Thus,  if  the  prbblem  were  to  find  the 
perpendicular  of  a  plane  triangle,  from  the  three  sides 
given,  it  will  be  much  better  to  substitute  for  one  of  the 
segments  of  the  base,  than  for  the  perpendicular,  though 
the  quantity  required ;  because  the  whole  bas^  being 
giveuj  the  other  segment  will  be  given,  or  expresseel,  ^y 
subtraction  only,  and  so  the  final  equation  come  out  a 
simple  one  ;  from  whence  the  segments  being  known,  the 
perpendicular  is  easily  found  by  common  arithmetic : 
whereas,  if  the  perpendicular  were  to  be  first  sought,  both 
the  segments  would  be  surd  quantities,  and  the  final  equa- 
tion an  ugly  quadratic  one. 

3°.  When,  in  any  problem,  there  are  two  lines  or 
quantities  alike  related  to  other  parts  of  the  figure,  or 
problem,  the  best  way  is  to  make  use  of  neither  of 
them,  but  to  substitute,  for  their  sum,  their  rectangle,  or 
the  sum  of  their  alternate  quotients,  or  for  some  line  or 
lines  in  the  figure,  to  which  they  have  both  the  same 
relation.  This  rule  is  exemplified  in  prob.  22,  23,  24, 
and  27. 

4°.  If  the  area,  or  the  perimeter  of  a  figure  be  givefi^ 
or  such  parts  thereof  as  have  but  a  remote  relation  to 
the  parts  required,  it  will  sometimes  be  of  use  to  assume 
another  figure  similar  to  the  proposed  one,  whereof  one 
side  is  unity,  or  some  other  known  quantity;  from 
whence  the  other  parts  of  this  figure,  by  the  known 
proportions  of  the  homologous  sides,  or  parts,  may  be 
found,  and  an  equation  obtained,  as  is  exemplified  in 
prot.  25  and  32. 

These  are  the  most  general  observations  I  have  been 
able  to  collect ;  which  I  shall  now  proceed  to  illustrate 
thy  proper  examples. 

PROBLEM  I. 

The  base  (^),  and  the  sum  of  the  hypothenuse  and  per- 
pendicular («),  of  a  right-angled  triangle^  ABC ^  being 
given  ;  to  find  the  perpendicnlar. 


256  The  Applicatmi  of  Algebra 

Let  the  perpendicular    BC  be  denoted   by  at;     then 

p   the  hypothenuse  AC  will 
^   be  expressed  by  a — x :  but 


{by  Euc.  47.  1.)  AB^  + 
BC2  =  AC^ ;  that  is,  b^ 
+  x^  •=■  c^  —  %ax_  +  XX  ; 

whence  x  = =  the 

2a 


B  perpendicular  required- 


PROBLEM  IL 

'The  diagonal  and  the  perimeter  of  a  rectangle^  A  BCD, 
being  given;  to  find  the  sides. 

Put  the  diagonal   BD  =  «,  half  the  perimeter   (DA 

+  AB)  =  b^  and  AB  = 
x\  then  will  AD  =  <^  — 
X ;  and,  therefore,  AB^  + 
KW  being  =  BD%  we 
have  x^  +  b^  —  2bx  +  x^ 
z=z  a^ ;  which,  solved,  gives 
\/2d^  —  b^  ^b 

PROBLEM  IIL 

The  area  of  a  right-angled  triangle  ABC,  and  the  sides 
of  a  rectangle^  Y.BDF  J  inscribed  therein^  being  given;  to 
deter ?nine  the  sides  of  the  triangle. 

Put  DF  =  Uy  DE  =  ^,   BC  =  Xy  and  the  measure 
Q       of   the    given  area  ABC 
=  d:  then,  by  similar  tri- 
angles,   we  shall    have  x 
—  b  (CF)  :  «(DF):  :x 

(BC)    :    AB  =  ^^^. 

rr.1  r  ax         ^       X 

Therefore  >  X   -  = 

X  —  b         2 

dy   and    consequently  ax'^ 

==  2dx  —  2hd,  or  ;^^  —  ~  =  —  — :    which,  solved, 
^  a  a 


to  Geometrical  Problems. 


257 


d 


^ives  X  z=,  —  ± 
a 


~   /7/Tr 


2bd 


,  from  whence  AB  and  AC 


will  likewise  be  known. 


PROBLEM  IV. 

Having  the  lengths  of  the  three  pcypendiculars  PF,  PG, 
PH,  drawn  from  a  certain  point  P  xvithin  an  equilateral- 
triangle  ABC,  to  the  three  sides  thereof;  from  thence  to 
determine  the  sides* 

L,tt  lines  be  drawn  from  P  to  the  three  angles  of  the 
triangle  ;     and   let  CD  be  perpendicular  to  AB  :     call 
PF   «;    PG  ^;    PH  c;    and 
AD    =    oc :      then    will    AC 
(=  AB)  =  2x,  and  CD  (== 
\/AC2  _  ^£)2j  _  V^3^^.  _ 

x\^  3  ;  and  consequently  the 
area  of  the  whole  triangle 
ABQ^(=:  CD  X  AD)  = 
xxs/ S.  But  this  triangle  is 
composed  of  the  three  trian- 
gles APB,  BPC,  and  APC  ; 
whereof  the  respective  areas  are 
ax^  bx^  and  ex.     Therefore  we  have  xxV  3  =  fi^  +  hx 


-f  ex  ;  and  from  thence,  by  division,  x  = 


PROBLEM  V. 


a  +  b^c 


Having  the  area  of  a  rectangle  DEFG  inscribed  in  a 
given  triangle  ABC ;  to  determine  the  sides  of  the  rectangle^ 

Let  CI  be  perpendi- 
cular to  AB,  cutting  DG 
in  H  ;  and  let  CI  =  a, 

AB  =  ^,  DG  =  X,  and  DX  H 

the    given   area    =  cc: 
then  it  will  be,  as  ^  : 

ax         ^,^- 
-  :    :«:---  =  CHj 

0 

which,  taken  from  CI, 


2og  The  Applicatmi  of  Algebra 

leaves  «•*—---  =  IH ;  and  this,  multiplied  by  x^  givers. 


ax 7~  :=z  cc  •=:  the  area  of  the  rectangle  ;  whence  we 

have  abx  —  ax^  =  hcc^  x^  ^^hx  ■=.  — 

^4         a  2^4         a 


have  abx  —  ax^  =  hcc.  x^  —  ^:v  = ,  :^  —  —  =  ± 


PROBLEM  VL 

Through  a  given  point  P,  xvithin  a  given  circle^  so  t& 
draw  a  right  line^  that  the  two  parts  thereof  PR,  PQ,  m- 
tercepted  between  that  point  and  the  circumference  of  the 
circle^  may  have  a  given  difference 

Let  the  diameter  APB  be  drawn;    and  let  AP  and 

BP,  the  two  parts  thereof 
(which  are  supposed  given) 
be  denoted  by  a  and  b ;  mak- 
ing PR  =  x^  and  PQ  =  x 
■\-  d{d  being  the  given  dif- 
ference). Then,  by  the  na- 
ture of  the  circle,  PQ  X  PR 
being  =  PA  x  PB,  we  have 
X  +  d  X  X  •=•  ah ^  or  XX  -^-^ 
dx-=^ah  \  whence  x  is  found 
=  V^F+Tdl—  \d. 


PROBLEM  VIL 

Fro7n  a  given  point  P,  without  a  given  circle^  so  to 
draw  a  right  line  PQ,  that  the  part  thereof  RQ,  inter- 
cepted by  the  circle^  shall  be  to  the  external  part  PR,  in  a 
given  ratio • 


to  Geometrical  Problems. 


259 


Through  the  center  O,  draw  PAB;    put  PA  =  tf, 
PB  =  /^,  PR  =  X,  and  let 
the  given  ratio    of   PR  to 
RQ    be    that    of   m  to  n ; 
then  it  will  be,  as  m  :  n  :  : 

—   =  RQ;    therefore 


X 


nx 


m 
PQ   =    PA     X     PB,    or 

^    nx  ,        , 

X  X  X  '\ =:  ab;    there- 

m 


fore    mx^  -f   nx^    =    maby 
and  X  =  \}- 


mab 


m  +  n 


PROBLEM  VIII. 

The  sum  of  the  two  sides  of  an  isosceles  triangle  ABC  Z?f- 
ing  equal  to  the  sum  of  the  base  and  perpendicular^  and  the 
area  of  the  triangle  being  given  ;  to  determine  the  sides. 

Put  the  semi-base  AD  =  x^  the  perpendicular  CD 
=  y,  and  the  given  area 
ABC  =1  a^  I  so  shall  xy  = 
a^,  and  2\^xx  -j-  ifi/  =  2x 
+  y  {by  EL  47,  1,  and  the 
conditions  of  the  problem). 
Now,  squaring  both  sides 
6f  the  last  equation,  we 
have  4*xx  +  4z/z/  =  A^xx  -f- 
4!xy  +  yy ;  whence  Syy  = 
4x2/,  and  consequently  y  = 

-— :  which  value,  substituted 

,  ^        AfXX 

in  the  former  equation,  gives  =  a^ ;    from  whence 


ha"         .      .— 


(=    -)  = 


faV  3  ; 


and  AC 


%m 


(=    V;vx    +    yt/ 


The  Application  of  Algebra 

Jsaa     ,     ^aa  hzSaa^ 


aV\  :=z\aV^. 


PROBLEM  IX. 


The  segments  of  the  base  AD  and  BD,  and  the  ratio  of 

the  sides  AC  and  BC,  of  any  plane  triangle  ABC  being 

given  ;  to  find  the  sides* 

Put  AD  =  «,  BD  ==  b, 

AC  =  .y  ;  and  let  the  given 

ratio  of  AC  to  BC  be  as  m 

nx 
to  n,    so   shall  BC  =  — • 

in 

But  AC2  _  AD^  (=  DCO 

=  BC^  —  BD%  that  is,  in 

Species,   ;c^  »—  a^ 


'b^. 


mm 
Hence  we  have  7r?x^ 


A 

and 


—  n^x^  -=.  rn^  X  aa-^  bb^ 


Jaa  — 

X  =  m\j ■ 

^  mm — ; 


bb 
nn 


PROBLEM  X. 


The  base  AB  (a),  the  perpendicular  CD  =  b^  and  the 
difference  (d)  of  the  sides  AC  —  BC,  of  any  plane  trian- 
gle ABC,  being  given  ;  to  determine  the  triangle.  (See  the 
preceding  figure,) 

Let  the  sum  of  the  sides  AC  +  BC  be  denoted  by  x\ 

dx 
then  {by  prop*  11,  sect.  18)  we  shall  have  a\  x  \  \  d  \  —  = 

the  difference  of  the  segments  of  the  base ;     therefore 

a  dx 

the    greater    segment    AD    will    be    =    -^    + 


aa    -f-    ^^ 


But    AD2    +    DC2 


2 
AC^ 


a"  •\'  2a^dx  -f  d^x'^ 
4aa 


+ 


_  x^  +  2dx  +  dd  ^ 

'  4 


2a 
that    is, 

whence 


to  Geometrical  Problems.  261 

a^  +  %a^dx  +  d^x^  +  4a^^^  =  a^x^  +  2aV;v  +  a^i^  ;  which, 

,      ,      .  \aa  4-  4bb  —  dd 

solved,  gives  x  =  ^V      aa-dd.     ' 


PROBLEM  XL 

The  base  AB,  the  sum  of  the  sides  AC  +  BC,  and  the 
length  of  the  line  CD  drawn  from  the  vertex  to  the  middle 
of  the  base^  being  given;  to  determine  the  triangle. 

Make  AD  (=  BD)  =  a,  DC  =  b,  AC  +  BC  =  c, 
and  AC  =  :v  :  so  shall  p 

BC  =  c  —  ^,      But  -^^ 

AC^  +  BC^  is  = 
2AD2  +  2DC2  (by 
El.  12,  2)  ;  that  is, 
o<^  +  c  —  x'Y  —  2a2 
+  2^2  ;  which,  by 
reduction,  becomes  x^ 

—    ex    :=z    d^    +   l^   — 

\c^  ;  whence  x  is  found  =  |c  ±  Vaa  + 

PROBLEM  XIL 

The  two  sides  AC,  BC,  and  the  line  CD  bisecting  the 
vertical  angle  of  a  plane  triangle  ABC,  being  given  ;  to 
find  the  base  AB. 

Call  AC  a;  BC  ^ ;  CD  c ;  and  AB.;\::  then 
a  +  b  '.  PC   :  I  a  '.  AD  = 


bi  DB  = 


and  a  +  b 

bx 


But  (by 


a  +  b 

El.  20,  3)   AC   X    CB  -, 

AD  X  DB  =  CD2,  that  is, 

-  abx^ 

ab  —  =r~  =  c^;    from 
axbj 

whence  x  wiU  be  found  =  a^b  .  ^* 


ab^ 


ab 


162 


The  Application  of  Algebra 


PROBLEM  XIIL 

The  perimeter  AB  +  BC  +  CA,  and  the  perpendicular 
BT^^  falling  from  the  right  angle  B,  to  the  hypothenuse  AC, 
being  given ;  to  determine  the  triangle* 

Let  BD  =  «,  AB  =  x,  BC  =  y,  AC  =  z,  and  AB 
+  BC  +  CA  =  ^:  then,  by  reason  of  the  similar  tri- 
angles ACB  and  ABD,  it  will  be,  as  2  : 2/  :  \x  \  a:,  and 

therefore  xy=iazi  more- 
over, x^+y'^  =2^  (by  Euc, 
47^  1),  and  x  -j-  y  -{-  z  =: 
b  (by  the  question^.  Trans- 
pose 2  in  the  last  equation^ 
and  square  both  sides,  and 
you  will  have  x"^  +  2xy  -j- 
y^=zb^  —  2bz  +  2^  ;  from 
which  take  x^  +  y"^  =.  2^, 
and  there  will  remain  2xy  •=.  b^  —  2^2  ;  but,  by  the 
first  equation,  2Arz/  is  =  %az  ;  therefore  2a2  "=.  IP'  —  2^2 


and  2  = 


2a  -f  2/^ ' 
a:  and  y  from  hence,  put 


whence  2  is  known. 
bb 


But  to  find 


=  c,  and  let  this  value 


2a  +  2^ 

of  2  be  substituted  in  the  two  foregoing  equations, 
X  +  y  ■=:  b  —  2,  and  zy  =  02,  and  they  will  become 
X  +  y  =.  b  —  c,  and  xy  :=:  ac  :  from  the  square  of  the 
former  of  which  subtract  the  quadruple  of  the  latter, 
so  shall  x^  —  2xy  +  y^  =z  b  —  c^  —  ^cic  ;  and  conse- 
quently X  —  z/  =  \  ^  —  cY  — "  4<2C.  This  equation  be- 
ing added  to,  and  subtracted  from  x  +  y  =  b  —  c, 

gives  2x  =z  b' —  c  +  \b  ■ 
_  \f^  _  cf  —  4ac. 


.  cl^  —  4aCy  and  22/  =  ^  —  c 


PROBLEM  XIV. 

Having  the  perimeter  of  a  right-angled  triangle  ABC, 
and  the  radius  DF  of  its  inscribed  circle;  to  determine 
all  the  sides  of  the  triangle* 


to  Geometrical  Problems. 


263 


From  the  center  D,  to  the  angular  points  A,  B,  C, 
and  the  points  of  contact  E,  F,  G,  let  lines  DA, 
DB,    DC,    DE,   DF,    DG  be  drawn;     making    DE, 


DF,  or  DG  —  a,   AB  =  x,  BC  c=  2/,  AC  = 


and 


It  is  evident  that  —  4- 
2    ^ 


x  +  y  +  z-b. 

ab 
2 

BDC,  and  ADC)    will  be   = 


ay         az 


or 


its  equal   —    (expressing  the  sum    of  the  areas  ADB, 


xt/ 
"2 


=    the   area  of  the 


whole  triangle  ABC ;  and  consequently  2x1/  =  2ab : 
moreover  (^z/  Euc.  47,  1),  x^  +  y^  =  z^  .  to  which  if 
2xy  =  2ab  be  added,  we  shall  have  x^  +  2xy  -f-  y^^  or 
.V  +  y1^  =  2^  +  2ab ;  but,  by  the  first  step,  x  +y']^  is 
=  b  —  2  I  2  =  ^2  __  2bz  +  7?  ;  therefore,  by  making 
these  two  values  of  x  +  y\^  equal  to  each  other,  we 
get  z^  +  2ab  ■=:  b^  —  2bz  +  z^  ;  whence  2a  =:  b  —  22, 
and  z  =  ^b  —  a.  But,  to  find  x  and  y,  from  hence, 
we  have  now  given  x  +  y  (=  b  —  2)  =  -^^  +  «,  and 
xy  =  ab :  the  former  of  these  equations,  multiplied  by 

x^  gives  :x^  +  xy  =:  —  +  ax  ;    from  which    the  latter 

xy  =  ab  being  subtracted,  we  have  x^  =  \bx  +  ax  - —  ab^ 

or  x^ — • — - —  X  X  =z — <  ai> :    whence,    by   completing. 


2G4  The  Application  of  Algebra 


.  .                2a  +  b  ±  V4a^  —  12ab  4-  b^ 
the  square,  ksrc,  x  = :  so 

that    the    three    sides    of   the    triangle    are,    ^b    —   a^ 
2a-\^b  +  V4a^  —  12ab  +  b^^ ^^^  2a  +  b—\^4a^~  1 2ab+b^ 
4  '  4 

Otherwise. 
The  right-angled  triangles  ADE,  ADG,  having  the 
sides  DE,  DG  equal  and  AD  common,  have  also 
AE  equal  to  AG :  and,  for  the  like  reason,  is  CE  = 
CF  J  and  consequently  AC  (AE  +  CE)  =  AG  +  CF. 
Whence  it  appears  that  the  hypothenuse  is  less  than  the 
sum  of  the  two  legs  AB  +  BC,  by  the  diameter  of 
the  inscribed  circle,  and  therefore  less  than  half  the  pe- 
rimeter by  the  semi-diameter  of  the  same  circle.  Hence 
we  have  AC  =  |(^  —  a,  and  AB  -|-  BC  =  |^  +  a.  Put, 
therefore,  |^  —  /2  =  c,  ^b  +  a  =  d^  and  half  the  dif- 
ference of  AB  and  BC  =  x  ;  then  will  AB  =z  d  +  x^  and 
BC  =zd—x;  and  consequently  2d^  +  2x^  (AB^  +  BC^) 
=  c^  ( AC^),  whence  x  is  found  =  V^  c^  —  d^  ;  therefore 
ABis^l^'  +  ^ic^  — ^,   andBC  =  |^— V'p'irS^- 

PROBLEM  XV. 

All  the  three  sides  of  a  triangle  ABC  being  give7i ;  to 
find  the  perpendicidary  the  segments  of  the  base^  the  area^ 
and  the  angles. 

Put  AC  =  (7,  AB  =  ^,  BC  =  c,  and  the  segment 
AD  =  X ;  then  BD  being  =  ^  —  x^  we  have  c^  — 
b  —  x'\  2  (=  CD2)  =  a^  —  x^,    that  is,    c^  —  b"^  -f-  2bx 


P  A  D       B 

x^  =  a"  —  x'^  '-,    whence  2bx  =  aa  +  bb  -^  cCj  and 


to  Ge07netrical  Problems*  '^^S 

2^ 


AC  +  AD  X  AC  —  AD  =  a  + 


aa  -^  bb  > —  cc 
2b 


aa    A-    bb  —  cc  2ab    +  aa    4-   bb   —  cc 

a    — ■' =    ' 1 V 

2b  2b         ^ 

2ab  —  aa  —  bb  4-  cc  _  «  -f  b^  —  c^        c^  —  a  —  b^ 

hence  CD  =  —  x  Ja  +  bf  —  c^  x  c^  —  ^^ZZ^Y ; 

and  the  area  ( )  = 

^         2         ^ 

\sj^T~bY  —  ?  X  c^  —  ^=Tp- 

But,  because  the  difference  of  the  squares  of  any  two 
lines,  or  numbers,  is  equal  to  a  rectangle  under  their 
sum  and  difference,  the  factor  a  +  bl^^  —  c^  will  be  = 
a  4-  b  -{-  c  X  a  '\-  b  —  c>,  and  the  remaining  factor 
c^  —  a  —  b']^:=zc  +  a*-^bxc*---a  +  b:  and  so  the  area 
will  be  likewise  truly  expressed  by 

i\ g  +  b  +  c  X  a  +  b  —  c  X  c  +  a-^b  X  c  —  a^ b 
la  -{-  b  -^  c      a  4-  b  —  c      c  4-  a* — *b      c- —  a  4-  b 

=  V-32j-^-V-^— i— ^— T^ 

=  /y  5  .s  —  c  .  s  —  b  .  s  —  a-yhy  making  s  =  — ^It . 

In  order  to  determine  the  angles,  which  yet  remain  to 
be  considered,  we  may  proceed  according  to  prop.  11, 
in  trigonometry,  by  first  finding  the  segments  of  the 
base:  but  there  is  another  proportion  frequently  used 
in  practice,  which  is  thus  derived :  let  B  A  be  produced 
to  F,  so  that  AF  may  be  =  AC  ;  and  then,  FC  being 
joined,  it  is  plain  that  the  angle  F  will  be  the  half  of 
the  angle  A  5   and  DF  ( =  AC  +  AD)  wili  be  given 

2  M 


266 


The  Application  of  Algebra 


{from  above)  = 


2b 


iff  —  ^  +  ^  -f  g       «  +  b  —  c 


~  c:    but  DF  (: 


^s    X    s   — 


)  is  to  DC 


C—\77s  — 


c  .  5  — -  ^  .  *  — .  a),  so  is  the  radius  to  the  tan- 


gent of  F  ;  and,  consequently,  s  X  s  —  c:**— ^X*  —  a 
:  :  sq.  rad.  :  sq.  tang,  of  F  ;  diat  is,  in  words,  as  the  rect- 
angle under  half  the  sum  of  the  three  sides,  and  the  excess 
of  that  half  sum  above  the  sid^  opposite  the  required 
angle,  is  to  the  rectangle  under  the  differences  between 
the  other  two  sides  and  the  said  half  sum,  so  is  the  square 
of  the  radius  to  the  square  of  the  tangent  of  half  the  an- 
gle sought. 

PROBLEM  XVI. 

Having  given  the  base  AB,  the  vertical  angle  ACB, 
and  the  right  line  CD,  which  bisects  the  vertical  angle ^  and 
is  terminated  by  the  base ;  to  find  the  sides  and  angles  of 
the  triangle. 

Conceive  a  circle  to  be  described  about  the  triangle, 

and  let  EG  be  a  diameter 
of  that  circle,  cutting  the 
base  AB  perpendicularly 
in  F ;  also,  from  the  cen- 
ter O  suppose  OA  and 
OB  to  be  drawn,  and  let 
CD  be  produced  to  E 
(for  it  will  meet  the  pe- 
riphery in  that  point,  be- 
cause the  angles  A  CD 
and  BCD,  being  equal, 
must  stand  upon  equal 
arches  EA  and  EB). 
Now,  because  the  angle  AOB  at  the  center,  standing 
upon  the  arch  AEB,  is  double  to  the  angle  ACB  at 
the  periphery,  standing  upon  the  same  arch  {Euc*  20.  3.), 


to  Geometrical  Problems •  267 

that  angle,  as  well  as  ACB,  is  given ;  and,  therefore, 
in  the  isosceles  triangle  AOB,  there  are  given  all  the 
angles  and  the  base  AB  ;  whence  AO  and  FO  will  be 
both  given,  by  plane  trigonometry,  and,  consequently, 
EF  (AO  —  FO)  and  EG  (  =  2AO),  Call,  therefore, 
EF  =  a,  EG  =  bj  CD  =  c,  and  DE  =  x;  and  sup- 
pose CG  to  be  drawn;  then,  the  angle  ECG  being  a 
right  one  (Euc.  31.  3.),  the  triangles  EDF  and  EGC 
will  be  similar ;  whence  x  :  a  :  :b  :  x  +  c;  therefore, 
by  multiplying  extremes  and  means,  we  have  x^  +  ex 
=  abj  and  consequently  x  =  \/ab  +  ^cc  —  ^c ;  from 
which  DF  (VED^  —  EF^),  half  the  difference  of  the 
segments  of  the  base,  will  be  found,  and  from  thence  all 
the  rest,  by  plane  trigonometry. 

Before  I  proceed  further  in  the  solution  of  problems,  it 
may  not  be  improper,  in  order  to  render  such  solutionis 
more  general,  to  say  something  here,  with  regard  tp  the 
geometrical  construction  of  the  three  forms  of  adiected 
quadratic  equations. 


mz. 


Cx^  +  aAT  =  be, 

<  x^  ^— a;c  =  be, 
{^ax-^x^'  =  be. 

Construction  of  the  first  and  second  forms. 

With  a  radius  equal  to  |a,  let  a  circle  OAF  be  de- 
scribed; in  which,  from  any  point  A  in  the  periphery, 
apply  AB  equal  to  b  —  c 
(b  being  supposed  greater 
than  c),  and  produce  the 
same  till  BC  becomes 
=  c  ;  and  from  C,  through  -p 
the.center  O,  draw  CDE  ^ 
cutting  the  periphery  in 
D  and  E  ;  then  will  the 
value  of  X  be  expounded 
by  CD,  in  the  first  case, 
and  by  CE,  in  the  second. 


268  The  Application  of  Algebra 

For,  since  (by  construction^  DE  is  =  a,  it  is  plain,  it' 
CD^be  called  x,  that  CE  will  be  x  +  a-,  but  if  CE  be 
called  a;,  then  CD  will  be  x  —  a\  but  (by  Euc.  37.  3.) 
CE  X  CD  =  AC  X  BC,  that  is,  ■%■  -f  a  x  x  (x^  +  ax) 
is  =  bcj  in  the  first  case  ;  and  x  X  x  '—  a  (x^  -^  ax)  =  bCy 
m  the  second :  which  two  are  the  very  equations  above 
exhibited. 

When  b  and  c  are  equal,  the  construction  will  be  ra- 
ther more  simple  ;  for,  AB  vanishing,  AC  will  then  coin- 
cide with  the  tangent  CF  ;  therefore,  if  a  right-angled 
triangle  OFC  be  constituted,  whose  two  legs,  OF  and 
FC,  are  equal,  respectively,  to  the  given  quantities  -Ja  and 
^,  then  will  CD  (  =  CD  —  OF)  be  the  true  value  of  x 
in  the  former  case,  and  CE  (  =  CD  +  OF)  its  true  value 
in  the  latter. 

Co7istruction  of  the  third  form. 

With  a  radius  equal  to  la,  let  a  circle  be  de- 
scribed (as  in  the  two  preceding  forms),  in  which  apply 

AB,  equal  to  the  sum  of  the 
two  given  quantities  ^  +  c, 
and  take  therein  AC  equal 
to  either  of  them ;  through 
C  draw  the  diameter  DCE  ; 
then  either  DC,  or  EC,  will 
be  the  root  of  the  equation. 

For,    the    whole    diameter 
ED  being  =  a,  it  is  evident 
B  that,    if   either    part    thereof 

(DC  or  EC)  be  denoted  by  x^  the  remaining  part  will  be 
a  —  x:  but  DC  X  EC  =  AC  x  CB  (Eiic.  35.  3.),  that 
is,  ax  — ♦  x^  =  bc^  as  was  to  be  shown. 

The  method  of  construction,  when  b  and  c  are  equal, 
is  no  ways  different ;  except  that  it  will  be  unnecessary  to 
describe  the  whole  circle;  for,  AC  being,  here,  perpen- 
dicular to  the  diameter  ED,  if  a  right-angled  triangle  OCA 
be  formed,  whose  hypothenuse  is  ^a,  and  one  of  its  legs 
(AC)  =  b,  it  is  evident  that  the  sum  (EC)  and  the  differ- 
ence (DC)  of  the  hypothenuse  and  the  other  leg  will  be 
the  two  values  of  x  required. 


to  Geometrical  Problems. 


269 


Note*  If  b  and  c  be  given  so  unequal,  that  b  -—*  c^  in  the 
'  two  first  fortns,  or  b  +^c,  in  the  last,  exceeds  (^)  the  whole 
diameter  ;  then,  instead  of  those  quantities,  you  may  make 
use  of  any  others,  as  ^b  and  2t:,  or  ^b  and  3Cj  whose  rect- 
angle or  product  is  the  same ;  or  you  may  find  a  mean 
proportional  between  them,  and  then  proceed  according 
to  the  latter  method. 


PROBLEM  XVII. 

The  bas-e  AB,  the  vertical  angle  ACB,  and  the  right  line 
CD,  drawn  from  the  vertical  angle  .^  to  bisect  the  base^  being 
given;  to  find  the  sides  and  perpendicular  • 

Suppose  a  circle  to  be  described  about  the  triangle; 
and  let  CQ  be  perpendicular  to  AB,  and  ED  equal  and 
parallel  to  CQ :  moreover,  from  the  center  F,  let  FA«, 
FB,  and  FC  be  drawn ;  also  !et  CE  be  drawn  (parallel 
to  AB).  Put  the 
sine  of  the  given 
angle  ACB,  to  the 
radius  1,  =  m,  its 
co-sine  =  n^  the 
xSemi-base  BD  =  ^, 
the  bisecting  line, 
CD  =  b,  ^d  the 
perpendicular  CQ 
(DE)  =  X',  then, 
since  (by  Euc.  20. 
3.)  the  angle  BFD 
is  equal  to  ACB,  it 
will  (by  plane  tri- 
gonometry) be,  as  m  (sine  of  BFD)  :  a  (DB)  :  t  n  (sine 

Tia 
of  PBF)  :  —  =  DF ;    and,  as  m  (sine  of  BFD)  :  a 

(DB)  :  :  1  (sine  of  BDF)  :  iL  =  the  radius  BF,  or 

m 

FC;  whence  EF  (ED -DF)  =;^-'if,  or  ^^^TTJ!?. 

m  m     . 

But  {by  Euc.  12.  2.)  DF»  +  FC«  +  2DF  X  FE  =  DC« ; 


270  The  Application  of  Algebra 

that  IS,  m  species,  — --  -j -| x =  6^^ 

nr         m^  m  m 

cP-         T?c^   .  'inax        ,,      i    ^      .  , 

or  — r H —  Ir :   but,  since  the  sum  of  the 

iir  rrr  m 

squares  of  the  sine  and  co-sine  of  any  angle  whatever 
is  equal  to  the  square  of  the  radius,  or,  in  the  present 
case,  w^  +  ^^*  =  1 ,  therefore  is  1  —  n^  =  wi^,  and  conse- 

^2  ^^2^2  ^2  ^^2 

quently  — ~  (or  —-  X  1  —  n^)  =  —  X  m^  =  a^  ; 

m^  rrr  nr  m^ 

2?2CIX 

whence  our  equation  becomes  a^  +  =  b^ ;    which, 

m 


ordered,  gives  x  ^  — — =  —  X 


n  AB 


m       DC  -^    DB  X  DC  —  DB        .         m 

—  X — tt; >  where  —  expresses  the 

n  AB  n 

tangent  of  the  angle  ACB  :  therefore,  in  any  plane  trian- 
gle, it  will  be,  as  the  base  is  to  the  sum  of  the  semi-base 
and  the  line  bisecting  the  base,  so  is  their  difference  to  a 
fourth  proportional ;  and  as  the  radius  is  to  the  tangent 
of  the  vertical  angle,  so  is  that  fourth  proportional  to  the 
perpendicular  height  of  the  triangle:  whence  the  sides 
are  easily  found. 

The  same  otherwise. 

Let  the  tangent  of  the  angle  ACB,  or  BFD,  be  re- 
presented by  j&,  and  the  rest  as  above  ;    then  it  will  be 

{by  trigonometry)  as  /?  :  1  (the  radius)  :  :  a  (BD)  :  — 

P 

==  DF  ;  therefore  FE  (DE  —  DF)  =  .r  ~  ~,  and  FC^ 

P 

(=  FB^  =  DB^  +  DF^)  =  ^2  ^  fl ;    and  consequently 
^  +  «'  +  5  +  7X^-y(I^F2  +  I'C'  +  2DFxFE) 
=  hb  (==  DC2),  that  is,  a^  +  —  -b^-,  whence  x  ^pX 
^~ — ,  the  same  as  before. 


to  Geometrical  Problems. 


271 


PROBLEM  XVIII. 

The  areaj  the  perimeter^  and  one  of  the  angles  of  any 
plane  triangle  ABC  being  given^  to  determine  the  triangle. 

Suppose  a  circle  to  be  inscribed  in  die  triangle,  touch- 
ing the  sides  thereof  in  the  points  D,  E,  and  F ;  ^Iso 
from  the  center  O,  suppose  OA,  OD,  OC,  OF,  OB, 
and  OE  to  be  drawn: 
and  upon  BC  let  fall  the 
perpendicular  AG ;  put- 
ting AB  +  BC  +  AC  = 
6,  the  given  area  =  a^, 
the  sine  of  the  angle  ACB 
(the  radius  being  1)  =  w, 
the  co-tangent  of  half  that 
angle  (or  the  tangent  of 
DOC)=n,^dAC  =  :c. 
Therefore,  since  the  area 
of  the  triangle  is  equal  to 
^ABxOE+iBCxOF 
+  ^AC  X  OD,  that  is,  equal  to  a  rectangle  under 
half  the  perimeter  and  the  radius  of  the  inscribed  circle, 


we  have  —  X  OE  =  aa  \ 


and  therefore  OE  = 


2aa 


But 


AD  being  =  AE,  and  BF  =  BE  ;  it  is  manifest  that  thei 
sum  of  the  sides,  CA  +  CB,  exceeds  the  base  AB,  h\f 
the  sum  of  the  two  equal  segments  CD  and  CF  ;  and  S0 
is  greater  than  half  the  perimeter  by  one  of  those  equf  il 
segments  CD;  that  is,  CA  +  CB  z=z  \b  +  CD:  but 
{by    trigonometry^    as    1    (radius)   :   n  (the  tangent    of 

DOC)  :   :  ^  (OD)  :  DC  =  ^;    whence  CA    + 


CB(=i^  +  CD)  =  l^  + 


2na^ 


which,  taken  from    (b) 


2na^ 


the  whole  perimeter,  leaves  ^b —  =  the  base  i  IB 


Make,  now,  ^b  -f- 


2na» 


=  c  ;  then  will  BC  =  c  =  a*  ;     also, 
(by  trigonometry^  it  will  be,  as  1  (radius)  :  m  (the    sine 


272 


The  Application  of  Algebra 


of  ACG)  :  :  X  (AC)  :  mx  =  AG ;   half  whereof,  mul- 
tiplied hy  c~x  (BC),  gives 


mcx 


»—  wr/v2 


•  =  a^,  the  area 
of  the    triangle:     from   whence  x  Comes  out  =  |c  ± 


4 


m 


PROBLEM  XIX. 

The  hypothenuse^  A,C^of  a  right-ajigled  triangle  ABC, 
aiyithe  side  of  the  inscribed  square  BEDF,  being  given; 
to  determine  the  other  tivo  sides  of  the  triangle. 

Let  DE,  or  DF  =  «r,  AC  =  b,  AB  ==  x,  and  BC  =  2/  ; 
p  then  it  will  be,  isls  x  :  y  -,  i  x 
^  — a  (AF):«(FD);  whence 
we  have  ax'=.yx  —  t/a,  and 
consequently  xy  •=.  ax  '\-  ay. 
Moreover,  xx  +  yy  =  bbi 
to  which  equation  let  the 
double  of  the  former  be  add- 
ed, and  there  arises  x^  +  2xy 


+  y^  :=.b^ 


+  2ax 
that  is,  X  +  yf  =  b^ 


+  2az/; 

^2a 

Xoc-\-y^orX'\-  ip^  —  2a  X  .^  +  ^  =  ^^  ;  where,  by  con- 
sidering :v  -f-  ?/  as  one  quantity,  and  completing  the 
square,  we  have  x  -f  y  Y  —  2a  X  ^  +  2/  -}-  a^  =  ^^  -|-  «2  . 
whence  x  -^-y  —  a  =  s/b^  +  a^,  and  ^  +  z/  =  Va^  -f-  b^ 
-f  <2  ;  which  put  =  c  :  then,  by  substituting  c  —  x  instead 
of  its  equal  (?/)  in  the  equation  xy  ■=:  ax  -^^  ay^  there 


will  arise  ex  • —  x^  =  ac 
V^c  —  ac,  and  y  =z-tc  - 


whence  x  will  be  found  =  |c  -{- 
•  V-^cc  —  ac. 


It  appears  from  hence  that  c,  or  its  equal  Vaa  +  ^^^ 
+  a,  cannot  be  less  than  4a,  and  therefore  ^^  not  less 
than  8a^ ;  because  the  quantity  ^cc  —  ac^  under  the 
radical  sign,  would  be  negative,  and  its  square  root  im- 
possible ;  it  being  known  that  all  squares,  whether  from 
positive  or  negative  roots,  are  positive  ;  so  that  there 
cannot    arise    any    such    thipgs    as    negative     squares, 


to  Geometrical  Problems. 


273 


F 


E 


imless  the  conditions  of  the  problem  under  consideration 
be  inconsistent  and  impossible.  And  this  may  be  demon- 
strated, from  geometrical  principles,  by  means  of  the  fol- 
lowing 

LEMMA. 

The  sum  of  the  squares  of  any  two  quantities  is  greater 
than  a  double  rectangle  under  those  quantitieSy  by  the  square 
of  the  difference  of  the  same  quantities. 

For,  let  the  greater  of  the  two  quantities  be  represented 
by  AB,  and  the  lesser  by  BC  (both  taken  in  the  same 
right  line).  Upon  AB  and  BC  let  the  squares  AK  and 
CE  be  constitut-  -^ 

ed;    take  AP  =       H  I  K 

BC,  and  complete 
the  rectangles  PH 
and  CF.  There- 
fore, because  AB 
=  AH,  and  AP 
=  BC,  it  is  plain 
that  PH  and  PD 
are  equal  to  two 
rectangles  under 
the  proposed  quan- 
tities AB  and  BC ;  but  these  two  rectangles  are  less 
than  the  two  squares  AK  and  CE,  which  make  up  the 
whole  figure,  by  the  square  FK,  that  is,  by  the  square  of 
PB,  the  difference  of  the  two  quantities  giv^n  ;  as  was  to 
be  proved. 

Now,  to  apply  this  to  the  matter  proposed,  let  there 
be  given  the  quadratic  equation  :>c^  -f  ^^  =  2ax^  or  x  =:  a 
±  s/aa  —  bb  :  then,  I  say,  this  equation  (and  consequently 
any  problem  wherein  it  arises)  will  be  impossible,  when 
Ofl  —  ^6  is  negative,  or  b  greater  than  a.  For,  since  b  is 
supposed  greater  than  a,  2bx  will  likewise  be  greater  than 
2ax  ;  but  2ax  is  given  :=z  xx  +  bb^  therefore  2bx  will  be 
greater  than  xx  -f-  bb^  that  is,  the  double  rectangle  of  two 
quantities  will  be  greater  than  the  sum  of  their  squares, 
tvhich  is  proved  to  be  impossible. 

2N 


274 


The  Application  of  Algebra 


PROBLEM  XX. 

The  base  AB  (a)  and  the  perpendicular  BC  {F)  of  a 
right-angled  triangle  ABC,  being  given;  it  is  proposed  to 
find  a  point  D  in  the  perpendicular^  such  that^  if  two  right 
lines  be  drawn  from  thence^  one  to  the  angular  point  A,  and 
the  other  (DE)  perpendicular  thereto^  the  triangles  DEC, 
ABD,  cut  off  by  those  lines  ^  shall  be  to  one  another  in  a 
given  ratio. 

Let  AB  be  produced  to  F,  so  that  the  angle  BFD  may 
be  equal  to  the  angle  BC  A  ;    putting  AC  =  c,  CD  =  a% 


T       ^ 

\e\ 

^y^U 

F            J 

3                    A 

and  the  given  ratio  of  the  triangle  DEC  to  ABD  as  ;w  to ;?. 

Then,  by  reason  of  the  similar  triangles   ABC,  DBF^ 

it  will  be,  a  (AB)  :  b  (BC)  x  i  b  —  x  (BD)  :  BF  == 

h^—hx         ,  .  „  h^  —  hx        a"  +  l^  —  bx 

;    whence  AF  =  a  -| = 

a  a  a 

=  ^Izi^  (because  a^  +  b^  -  c^\      Also,  as  ADE  is  a 
a        ^  ^ 

right  angle,  the  angles  FAD,  EDC  will  be  equal ;  there- 
fore, the  angles  C  and  F  being  equal  {by  con.)^  the  tri- 
angles AFD,  DCE  must  be  similar ;  and  consequendy 
^p     c^  — ^>a1\  .  ^^o^..<,^..  AF  X  BD  ,b—xxc^~bx 


-):CD2(^2).. 


•(- 


2a 


). 


b  — —  X  X  ax^ 

the  area  of  the  triangle  AFD  :  ( ==-),  the  area 

2X6-2  —  bx 

of  the  triangle  DEC  :    wherefore,  the  area  of  the  tri- 


to  Geometrical  Problems.  275 


1    ATiTM.  •      BD  X  AB       b  —  X  X  ct         1,111 
angle  ABD  being ,  or ,  we  shall  have 


which,  reduced,  gives  x  =  vf —  H —  ■ 


b  —  X  X  a-^    b  —  X  X  a  . 

mini  ;  -■  :  • {by  the  question)  ;  and 

^  X  c  ' 

-  9  ,    -^ «    ,    mbx       mc^ 

consequently  nx^  =:  m  x  c^  —  f^^^  or  x^  +  =  — ^ 

•^  71  n 

mb 

n     '    4n^        2n 
The  geometrical  construction  of  this  problem,  from 

TYtbx  j^f^^ 

the  equation  x'^  4 =  — ,  may  be  as  follows;      In 

71  n 

CB  let  there  be  taken  CH  :  CB  :  :  lyi  i  n,  and  let  HK 
be  drawn  parallel  to  BA  ;  then  CH  being  =  — ,  and  CK 

71 

Vie 
=  — ,  our  equation  will  be  changed  to  x^  +  x  X  CH 

=  AC  X  CK,  or  to  CD  x  CD  +  CH  =  AC  x  CK. 

Upon  CH  as  a  diameter  let  the  circle  CTHQ  be  describ- 
ed, in  which  inscribe  CG  =  AK  ;  and  in  CG,  produced, 
take  CS  =  CA  ;  and  from  S,  through  the  center  O, 
draw  the  right  line  STOQ,  cutting  the  circumference 
in  T  and  Q,  and  make  CD  =  ST ;  then  will  D  be  the 
point  required.  For  CG  being  =  AK,  and  CS  =  CA  ; 
therefore  will  AC  x  CK  =  CS  x  GS  =  ST  x  SQ  {Euc. 
37.  3.)  =  ST  X  ST  +  TQ  =  CD  X  CD  +  CH  5  the  very 
same  as  above. 


PROBLEM  XXI. 

Having  the  perimeter  of  a  right-angled  triangle  ABC, 
and  three  perpendiculars  DE,  DF,  and  DG^  falling  front 
a  point  within  the  triangle  upon  the  three  sides  thereof;  t<i 
determine  the  sides. 

Suppose  DA,  DB,  and  DC  to  be  drawn;  and  let  DE 
=  a,  DF  =  b,  DG  =  c,  AB  =  a;,  BC  ==  y,  AC  =  z. 


sre 


The  Application  of  Algebra 


and  the  perimeter,  AB  +  BC  +  AC,  =  p ;    then,  the 

Q  area  of  ADB  being  ex^ 

that 


that 


pounded 

by 

ax 

"a' 

of  BDC 

by 

of  ADC 

by 

cz 

and 


that  of  the  whole,  ABC, 
by  — ^ ;  we  therefore  have 


,  or  ax  +  by  +  cz  :=.  ocy  I 


more- 


ax       'by        cz        xy 

over,  we  have  x^  -^  y^  -=  z^^  and  x  +  y  +  z  =  p^  by 
the  conditions  of  the  problem.  Let  z  be  transposed  in 
the  last  equation,  and  both  sides  squared,  so  shall  .x^  -f  2xy 
+  y^  ■=  p^  —  2pz  -h  2^,  from  which,  if  x^  -f-  y^ 
s=  z^  be  subtracted,  there  will  remain  2xy  =  /^^  — 
2pz  =  2ax  +  2by  4-  2cz  (by  the  jirst  equation)  :  whence 
ax  +  by-\'C'\'pXz-=.  \pp :  from  this  last  equa- 
tion subtract  a  times  a:  +  ?/  4-  2  =  /?,  and  there  will 
remain  by  —  ay  +  p  -^  c  —  a  X  z  =  ^p^  > — >  ap ; 
also,  if,  from  the  same  equation,  b  times  x  +  y  +  z 
=  /)  be  subtracted,  there  will  remain  ax  —  bx  + 
p  +  c  —  b  X  z  =  ^p^  —  bp  ;  which  two  last  equations, 
by  putting  d  z=z  b  —  a^  e  =  p  +  c  —  ci',  f  =  I  ^^  —  ctpy 
g  z^  p  +  c  —  ^,  and  h  =  ^p^  —  bp^  will  stand  thus  : 
dy  +  ez   =  fj    and  —  dx  +  gz  =  h ;     whence  y  = 

/*—  ez         ,  ^z  —  h 

*J . — ^  and  X  =  ^5 . — . 

a  a 


Let  these  values  of  x  and  y 


be  substituted    in    :\^  -f 

p  —  2€fz  -^  (^Z^        g^^^  — 


2^,    and  we  shall    have 


^rhz4-h^ 


//2 


d^ 


=  2^,  or  e^  -!-  g^  —  d^ 


XZ^  —  2/4-  ^gh  X  z  =  — /2  —  A2  :   put  e^  +  ^2  _  ^2 
^  k,  ef  ■\-  gh  =z  I,  and  f^-{-h^z=zm;    so  shall  J^z^  ~ 

2/2  =  —  m;    whence  2^  —  --~  =  ^ ~,  and  2  =  7- 

k  k  k 


to  Geometrical  Problems. 


^77 


,      //2         m        I 

*  V  IS  ~  T  =  - 


±  V/^  —  km 


from  which  x  (^z:z 


^' 


— )  and  y  (  =—7 — )  will  also  be  known. 

a  d 

If  a,  b^  and  c  be  all  equal  to  each  other,  the  point  D 
will  be  the  center,  and  each  of  the  given  perpendiculars 
a  radius  of  the  inscribed  circle  ;  and  the  value  of  2,  in  this 
case,  will  be  barely  equal  Xo\p*-^a\  for  the  equation,  by 
—  ay  +  p  +  c  —  a  X'Zz=.  \p^  —  ap^  above  found,  here 
becomes  pz  =  \p^  —  ap. 

But,  if  only  a  and  b  (or  DE  and  DF)  be  equal,  then 
the  equation  will  become  /?  +  c  —  aX2:  =  \p'^  —  ap  \  and 

therefore  z  =  -^ ^  ;    in  which,  if  c  be  taken  =  0* 

p  ^c  —  a  ^ 

2  will  be  =  1^- -^ ;  where  a  is  the  side  of  the  inscrib- 

p  —  a 

ed  square. 


PROBLEM  XXir. 

The  perpendicular  CD,  the  difference  of  the  sides^  AD 

—  BD,  and  the  vertical  angle  D,  of  any  plane  triangle 
ABD,  being  given;  to  determine  the  sides* 

From  B,upon  AD  (produced,  if  need  be),  let  fall  the 
perpendicular  BE  :  let  the  sine  of  the  angle  BDE  =  5, 
its  co-sine  =  c  (the  radius  being  unity)  \  also  let  the 
perpendicular  CD  =  /?,  the 
lesser  side  BD  =  .v,  and  the 
greater  DA  =  a:  -f  ^;  then 
{by  prop.  9,  in  trigonometry^ 
2&  \  ',  s  \  \  X  \  sx  z=z  BE  ; 
and  as  1  :  c  :  :  a;  :  c^  = 
ED.  Now  AB2,  being  = 
AD^  +  DB2_ADx2DE 
{Euc.   13.  2.),  will  be  ex- 

pounded  by  x  '\-  ([f  ^  x^    k  c  R 

—  X  •\'  d  X  2cx^  or  2x^  -f- 

2dx  +  d^  .^  2cx^  —  2cdx ;    whence,  by  reason  of  the 


278  The  Application  of  Algebra 

similar  triangles  ABE  and  ADC,  it  will  be,  as  Sat^  ^ 
^dx  ^  d^  —  2cx^  —  2cdx  (AB^)  :  s^x^  (BE^)  :  :  cc^ 
+  2xd  +  dd  (AD2)  \  p^  (DC2);  and,  consequently, 
by  multiplying  extremes  and  means,  s^x^  -f  2s^dx^  + 
^d^x^  =  2^2  .2  ^  2^2^;^  ^  ^2^2  _  2/^2c;c2  —  2p^cdx  ;  from 
whence,    by  transposition  and  division,    we  have  x"^  + 

X  —  ^— ^  =  0.  Which  equation  answering  the  condi- 
tions of  the  second  case  of  biquadratics,  explained  at 
p.  154,  we  shall  therefore  have  x^  +  dx  -^^ 


p^c  —  p^ 


12 


//,2^2     "i^x  —  p^  r 

\'— J-    +  ^ -^ — i- ;    and  consequently  x  =  —  |^  + 


Supposing  ^,  c,  and  j&  to  be  the  same  as  before,  put 
half  the  given  difference  of  the  sides  =  a,  and  half  their 
sum  =  X ;  then  the  greater  side  AD  will  be  =  a:  -f  a;, 
and  the  lesser  BD  ■=  x —  a  ;  wherefore  (by  trigonometry) 
1  \  s  I  I  X  —  a  \  s  X  ^  —  ^  =  BE  ;  and  1  :  c  :  :  .r  — -  a 
:  c  X  X  —  a  =  DE  :  but  AB^  is  =  AD^  -f  DB^  —  2DE 
X  AD  z=zx  4-al2  +  X  —  oY  —  2c  X  oc  —  ax  x  +  a=i 
2x^  4-  "^a^  —  2cr^  +  2ca^  ;  whence,  by  reason  of  the 
similar  triangles  ABE,  ADC,  it  will  be  2x^  -f  2^^  ~ 
2c  .2  +  2ca2  (AB2)  :  s^  x  x  —  a]^  (BE^)  :  :  x  +  a]^ 
(AD^)  :  /?2  (DC^);  and  consequently  s^  x  x  —  af  X 
X  4-  ^1^  =  S^Y^  ^  2(^2  —  2cx^  -f  2C(22  X  p^,  or  ^^^  4  _ 
252^^x2  ^  ^2^4  ^  2/?2;c2  —  2cp^x^  +  2/?2flr2  ^  2cp^a^ ; 
whence,  by  transposition  and  division,  x"^  —  2a^x^  — 

2^  +  !f^  =  ?^!  +  ^    ^  a^.      Substitute 


A" 


to  Geometrical  Problems.  279 


/=  fl2  +Pl-%,  and  5-  =  «^  X  ^UrJfL  -  a^.  then 
the  equation  will  stand  thus,  x^  —  2fx^  =  g  :    whence  i;^ 

is  found  =  Sf±  V/2  -f  ^. 

If,  instead  of  the  diiference,  the  sum  of  the  sides  had 
been  given,  in  order  to  find  the  difference,  the  naethod  of 
operation  would  have  been  the  very  same ;  only,  instead  of 
finding  the  value  of  x  in  terms  of  a^  by  means  of  the  equa- 
tion s^x^  —  2s^a^x^  +  s'^a^  =  2fx^  —  2cp^x^  +  2p^d^  + 
2cp^a^^  that  of  a  must  have  been  found,  in  terms  of  ►r,  from 
the  same  equation. 

PROBLEM  XXIII. 

Having  one  leg  AB  of  a  right-angled  triangle  ABC  ^ 
to  find  the  other  leg  BC,  such  that  the  rectangle  under  their 
difference  {^Q,  —  AB)  and  the  hypothenuse  AC  may  be 
equal  to  the  area  of  the  triangle. 

Put  AB  =  a,  and  BC  =  ^t- ;  so  shall  AC  =.Vaa  +  xx\ 


and  —-  =  .V  — a  .  Vaa  +  xx^  by  the  conditions  of  the 

problem.  By  squaring  both  sides 
of  this  equation,  we  have  ^a^x^  = 
;^  —  2ax  +  a^  x  aa  -f  xx\  in 
which  the  quantities  x  and  a  being 
concerned  exactly  alike,  the  solu- 
tion will  therefore  be  brought  out 
from  the  general  method  for  ex- 
tracting the  roots  of  these  kinds  of 
equations  ^delivered  at  p.  156): 
according  to  which,  having  di- 
vided the  whole  by  c^xP^^  we  get  —  =  -21  ,— ,  2  +  -^ 
4  a  X 

X  —  H ;   which,  by  making  2  =  -^  +  — ,  will  be 

X         a  ax 

reduced  down  to  l  =  ^  ~  2x2,  or  z^  —  2z  =  1 : 

-^hence  z  is  given  =;  1  +  V  f .     But  since  —  +  —  =:  2, 

a    '    X         ' 


280  The  Application  of  Algebra 

we  have  jc*  —  azx  =  . —  a^  \    and  therefore  ;v  =  —  + 
2   ^ 

y  — «2  =  _X2:  +  V22  —  4:    which,  by  sub- 

stituting  the  value  of  2,  becomes  :v  =  —  X 

At 

i  +  Vf  +  Vv'i'— J. 

PROBLEM  XXIV. 

To  ^mw  a  right  line  DF  yro;?z  one  angle  D  of  a  given 
rhombus  ABCD,  so  that  the  part  thereof  ¥G^  intercepted  by 
one  of  the  sides  including  the  opposite  angle  and  the  other 
side  produced^  may  be  of  a  given  length. 

Let  DE  be  perpendicular  to  AB  ;  and  let  AB  (  =  AD) 
C  =  «,  AE  =  b,  FG 

=  c,  and  AF  =  x : 
then  DF2  (=  AF^ 
+AD2— 2AExAF) 

:=z  XX  +  aa  —  2bx  ; 
and,  b}^  similar  tri- 

_ -^ —^     angles,  xx  -^  aa  • — 

A  E  B  F     2bx(DF^):xx(AF^) 

:  cc  fFG2)   :   x  — n  o']^    TBF^)  ;      and    consequently 
^x  4-  aa  —  2bx  X  xx  —  2ax  4-  aa  z=:  ccxx.     Make  ma  = 
^,  and  na  =  c  ;  so  sh'i;l  our  equation  becon>e 
XX  +  aa  —  2max  x  ^^  —  2a    -J-aa^  n^a^x^  :  which^  di* 

X         fi                       X         a 
vided  by  a^x^.  jrives f- 2w  x 1 2  = 

•^  °  a        AT  ax 


J 


XX 


X        a  

n'^  :  this,  by  making  2  = f >  becomes  z  —  2m  X 

Z  —  2  =  72*  :  therefore  z^  —  2m  -f-  2  X  2  ==  n^  —  4w, 
and  2  =  1+  yyz  4-  sjn^   -J-  1  —  wz")^  = 

^  +  ^4-  Vc^   +  a  -  /^?     by  restoring  the    values 

a  " 

m  and  72.      From  whence  the  value  of  x  will  be  also  J| 


to  Geometrical  Problems., 


^l 


known  \    for 1 being  =  ^,  we  have,  by  recluc- 

a  X 

tion,    x^  —  azx  =  —  aw,     and   therefore    a:  =  —    x 


2  +  V'2^  4. 


PROBLEM  XXV. 


TA^  diagonals  AC,  BD,  and  all  the  cmgles  DAB,  ABC, 
BCD,  and  CD  A,  of  a  trapezium  ABCD,  being  giv^n; 
to  determine  the  sides. 

Let  PQRS  be  another  trapezium  similar  to  ABCD, 
whose  side  PQ  is  unity  j  and  let  QP  and  RS  be  produced 
till  they  meet  in  T :  also  let  PR  and  QS  be  drawn, 
and  make  Rz;  and  Sw  perpendicular  to  TQ.      Let  the 


(natural)  sine  of  the  given  angle  STP,  to  the  radius  1, 
be  put  =  wz ;  that  of  TSP,  or  PSR,  =  n ;  that  of 
TRQ  =/; ;  the  CO  sine  of  SPQ  =  r  ;  that  of  RQv  =  s  ; 
AC  =  a;  BD  :::::  ^  ;  and  PT  =  X.     Then  (bij  plane  tm- 

gonometry)  n  :  m  -.  -.  x  x  V^  =  — -,    and  1   :  ^  f  PS) 

71  n 

.iiriFwzzz^IHf^;  whence  (by  Euc.  13.  2.)  QS^  (=  QP^ 

+  ps^  ^  2PQ  X  Pzt.)  =  1 +^!:fi!  ~  £!:^. 

nn  n 

Again  {by  trigonometry)^  p  :   m   :   \    1  +  x  (TQ)   : 


QR  = 


•  mx 


and  1  :  s 


m  4-  mx 


(QR)  :  Qy 


P       '  P 

7ns    -|-    TUSX 

T And  therefore  PR^  ( ~  PQ^  +  QR2  _ 

20 


282  The  Application  of  Algebra 

2PQ  X  Qt>)  =  1  +  ^ElHI  _  ^J!1L±3^.     But 

Z'/  P 

because  of  the  similar  figures   ABCD,  PQRS,  it  will 
be,    AO    :    Bpg   :   :   PR*   :    QS%    that  is,    a^  :   b^  :   : 

m  -i-  /.,^j  2??2^  4-  2m,sx  m^x^         2rmx 

1    -f-   , _    ;     1    -J-    —   — -    • 

pp  p  nn  71 

-  -       «        a^/  V         2a^rmx         ...        Z>^/?i^ 

and  consequently  a^  ^ *  z=z  b^  -] 

rm  n  pp 

2b^mP-x^         b^m^x^          2b^ms       2l^msx 
4 1 -—  — • :  whence,  writ- 

PP  PP  P  P 

^      a^m^        b'^m^  bbms         bbmP'         aarm         , 

mc:  /  = ,  9"  = ,  and 

^  ^  nn  pp  p  pp  n    ' 

uTi^       ^i/bbtyis 
h-=.h^  —  c^  A ,  we  have  fx^  +  2^-;^  =  h  : 

pp  p      ' J         -r     ^ 

which,  solved,  gives  x  =  V -77  +  -%  —  -^  •   from  whence 

SQ  will  also  be  known  :  and  again,  by  reason  of  the  simi- 
lar figures,  it  will  be  as  QS  :  QP  (unity)  :  :  BD  :  AB  ; 
which,  therefore,  is  likewise  known :  from  whence  the 
rest  of  the  sides  BC,  CD,  and  DA  will  all  be  found  by 
plane  trigonometry. 

The  last  problem  is  indeterminate  in  that  particular  case, 
where  the  trapezium  may  be  inscribed  in  a  circle,  or  where 
the  sum  of  the  two  opposite  angles  is  equal  to  two  right 
ones  ;  for  then  there  can  but  one  diagonal  be  given  in  the 
question,  because  the  value  of  the  other  depends  entirely 
upon  that. 

PROBLEM  XXVI. 

Supposing  BOD  to  be  a  quadrant  of  a  given  circle  ;  to 
fnd  the  scmi'diameter  CE,  or  CL,  of  the  circle  CEGL, 
inscribed  therein ;  and  likewise  the  semi-diameter  of  the 
little  circle  n¥mV^  touching  both  the  other  circles  DLB, 
LME,  and  also  the  right  line  OB. 

Let  BQ,  P/z,  and  CE  be  perpendicular  to  BO  ;  join 
C,  7iy  and  O,  n ;  and  draw  OC  meeting  BQ  in  Q, 
and    nr    parallel  to  BO,    meeting  CE  in  r :     put  OB 


io  Geometrical  Problems. 


283 


(  =  BQ)  =  1,  OQ  (=  V2,  /^j/  Euc.  47.  1.)  =  h,  and 
Pn  C  =  nni)  =  a:.  Then,  by  reason  of  the  similar  tri-^ 
angles   OBQ,  OEC,    it    will  be,  OQ  :   BQ  :   ;   OC   : 


B    P  E 

CE  :    whence,  hy  composition^  OQ  +  BQ 
(OC  +  CE)  :  CE  ;    that  is,  ^  +  1  :  1 


0 

BQ 

;  1  : 


:  :  QL 
CE  = 


=  i^-^l   =  ^2 


X   b—V   ~  ^2  —  1 

— -*  1  ;    which  let  be  denoted  by  a,  then  we  shall  have 

Ctz  +  Cr  =  2<2,  and  Cn  —  Cr  =  2^; ;    and  therefore  nr 

(  =  V^^^  +  Cr  X  C7i  — .  Cr)  =  2>/a;c.  Moreover,  On 
+  P/z  being  =  1 ,  and  On  —  P/i  =  1  —  *2.x  \  thence  will 
OP^  V  1  —  2:v ;  which,  also,  being  =  PE  +  OE 
(2^6  a:  +  «),  we  therefore  have  V  1  — •  2:v  =  "ilS^ax 
+  <^  ;  whereof  both  sides  being  squared,  there  arises  1  — 
2a;  =  A^ax  -f-  Aas/ax  +  a^,  or  1  —  c^  —  2a:  —  Aax  = 
'\a>/ax ;  which,  because  1  —  aa  is  2^7,  will  be  «  — 
1  +  2a  X  -^^  =  "la's/ ax  :  this,  squared,  gives  a^  —  1  -f-  2a 
X  2aa"  +  1  +  2a1^  X  ^^  =  ^a^-^  j  whence  1  +  2a}^  x  x- 
—  1  +  2a  X  2a;v  —  4«^.^  =  —  ^« ;    "which,  by  writing 


2B4  The  Application  of  Algebra 

b  —  1   instead  of  its  equal  a^  becomes  2^  • —  \\^  X  ^^  — 

^ Yb 9 

7i)  —  9  X  2x  z=  2b  —  3  ;    therefore  x^  —  —  X 

2b  —  1 Y 

2x  =  =z==z-=r- ;  from  whence  x  is  found  = 
2u  —  i7 

II 

7  b  — 9  '^  sJ'yJj  —  ;f.  x  ^o  —.  i]^  +  7b  —  ^Y 


'tzz2^..JE£'     ^7b-<q^  ^ 


"Zb 


'=^' 


' :    which,  by  wnjt- 


'zb—l\' 


ing  V  2  for  b^  becomes    ^- ""  ___       ' 5^    — 

7V2  —  9  t'-  6V/2  IP  8     ,        .  -         13^^2  —  17 

-:===— ,  that  IS,  equal  to — , 

2V2^_l]  2V2  — 1> 

V  2  —  1  ,  .  ,    ,         .       , 

or  to  — : rzr^  ;    which  last  is  the  root  required,  the 

2V'  2  —  1  I 

other  being  manifestly  too  large  :    but  this  value  will  be 

"49" 


reduced  to  ^ .      Therefore  OP  (  =  V  1  —  2x) 


*,       .  /5I  — 10.2       5V'^  — 1        ^^  _ 

IS  given  =  y = =  7Vn  ;  and  con- 

^  4-9  7 

sequently  BH  =  ^-^Q  5  from  whence  we  have  the  follow- 
ing construction. 

In  the  tangent  BQ  take  BH  =  -|.BO  ;  draw  HO,  cut- 
ting the  circumference  BDL  in  F,  and  make  the  angle 
OFP  =  iOHB,  and  draw  PN  parallel  to  BQ,  meeting 
OH  in  72,  the  center  of  the  lesser  circle  required. 


SCHOLIUM. 

in  the  preceding  solution  it  was  required,  not  only  to 
extract  the  square  root  of  the  radical  quantities  136  -^ 


to  Geometrical  Problems.  285 

96  y  2  and  51  —  lOv/2,  but  likewise  to  take  away  the 
radical  quantity  from    the    denominator  of  the  fraction 

v^T ■'  1 

— -^  ,  and  confine  it,  wholly,  to  the  numerator: 

2V  2  —  1  I 

all  which  being  somewhat  difficult  (and,  for  that  reason, 
omitted  in  the  introduction,  as  too  discouraging  to  a  young 
beginner),  I  shall  therefore  take  the  opportunity  to  ex- 
plain here  the  manner  of  proceeding  in  such  like  cases, 
when  they  happen  to  occur. 

First,  then,  with  regard  to  the  extraction  of  roots  out  of 
radical  quantities,  let  there  be  proposed  A  ±  v  B^  A  be- 
ing the  rational,  and  VB  the  irrational  part  thereof; 
and  let  the  root  required  be  represented  by  y/~±  vTT- 
the  square  of  which  will  be  x  +  y  ±  2v'^^  or  x  +  y 
±  V4:xy  ;  therefore  we  have  x  +  y  ±  V4xy  =  A  rh  VBI 
Let  the  irrational,  as  well  as  the  rational  parts  of  these 
two  equal  quantities  be  now  compared  together ;  so 
shall  X  +  y  z=:  Aj  and  V4:xy  =  VB  :  from  the  square 
of  the  former  of  which  equations  subtract  that  of  the 

latter  ;   whence  will  be  had  xx  —  2xy  +  yy  =  A^ B  ; 

and,  by  taking  the  square  root,  x  —  y  =  v  A^  b"; 

which  added  to,  and  subtracted  from  x  +  y  ~  A,  ^c. 

A  +  VA^  —  B        ,          A  — VA^TZb" 
gives  X  = ^ ,  and  y  = -_ r. 

In  the  first  of  the  two  cases  above  specified,  the  quan- 
tity whose  square  root  is  to  be  extracted  being  136  

96V  2,  we  have  A  =  136,  and  B  =  18432  (=  96]^ 
X  2}  ;    whence  we  have  x  {—  A  +  V  A^  —  B^  _  ^^^ 

A      a/  A  2  P 

and  y  {=: )   =  64;    and  consequently 

\/x~V^  =z  V72  —  V64  =  6V2"—  8,  the  required 
square  root  of  136  — -  96V  2.  After  the  very  same  man- 
ner, die  square  root  of  the  other  radical  quantity  51  

10V2,  or  51  —  V200  will  be  found  to  be  5^2—  1  : 


286  The  Application  of  Algebra 

for,  A  being  here  =  51,  and  B  =  200,  we  have  .v  =  50^ 
and  2/  =  1  ;  and  consequently  V  x  —  V^ «/  =  5V  2  —  1. 

What  has  been  said,  thus  far,  relates  to  the  extraction 
of  the  square  root  only  ;  but  the  same  method  is  easily 
extended  to  the  cube,  biquadratic,  or  any  other  root. 
Let  us  take  an  instance  thereof  in  the  cube  root ;  where 
we  will  suppose  the  given  quantity,  out  of  which  the 
root  is  to  be  extracted,  to  be  represented  by  A  ±  VB, 
as  before.  Then,  if  the  rational  part  of  the  root  be  de- 
noted by  x^  and  the  irrational  part  by  V  y  ;  the  root  itself 
will  be  expressed  by  ^  ±  V  y  ;  and  its  cube  by  x^  ± 
3x^  V  y -^  3xfj  ±  y  V  1/ :  from  whence,  by  proceeding  as 
in  the  extraction  of  the  square  root,  we  have  x^  +  2xy 
*:  A,  and  3x^  \/  y  +  y  V  y  =:  VB,  Let  the  sum  and  the 
difference  of  these  two  equations  be  taken,  and  there 
will  come  out  x^  -f  3x^  ^  y  +  ^^y  +  ^/  V  ^  =  A  +  v'B, 
and  x^  —  3x^  V~y  -f  3xy  —  y  V'y  =  A  —  VB  ;  where- 
of the  cube  root  being  extracted  on  both  sides,  we  thence 

have  x  +\/y  =  A  -fv'B]^, and  x^^\/y  =  A  — v  B]^  - 
let  the  two  last  equations  be  added  together,  and  the  sum 

be  divided  by  2  :  so  shall  x  =  — i-!^ — ! — ^ — *■    ; 

and,    by  multiplying  the  same    equations    together,    wc 

get  ^:^  ' — '2/  =  A^  — ;>  B")  s",  and  consequently  y  :=i  x^  — ^ 

A^  —  B"J^  ;  whence  y  is  likewise  known. 

Universally^  let  the  index  of  the  root  to  be  extracted 
be  denoted  by  /2,  and  let  the  root  itself  be  represented 

by  x±.s/  y  (as  above).      Then  this  expression  raised  to 

11      -  1 
the  nih  power,  will  be  x""  ±  nx  "~V  y  +  n  X  — r —  ^"^1/ 

±  n  X   — -—     X  : x^-^y  V  2/,  &c.  from  whence, 

2  3 

still    following    the    same    method,    we    shall   here    get 

72  —  1 

x^  ^nx  ^''^"^y-i  Sec.  =  A,  and 


to  Geometrical  Problems.  287 

v2.Y«-V"^  +  n  X  ^~     X  ^  x^-^y  V"y,  &c.  =  VB  : 

let,  therefore,  the  root  of  the  sum,  and  also  of  the  dif- 
ference of  these  two  last  equations,  be  taken,  and  you 

will  have  a:-  +  V  z/  =  A  +  y/B^  ^,  and  x  en  V  if  =z 
A  —  v/  B  I  '^ ;  which  two  equations  being  added  to- 
gether, X  will  be  found  =  i— ^^ ! ^ — L 

A  -f  v'B1«  _^        A^  —  B]"^  ,    .,    , 

::=: -1 — 1-  ± y  *.     and,  ir  the  same 

^  2  X  A  +  v^l~' 

equations     be     multiplied     together,      you     will     have 

L  ,L 

x^  CO  1/  =  A2  —  B  J  «  ;  whence  ^2/  =  .r^  ±  A^  —  B  |  «  :  the 
use  of  which  conclusions  will  appear  by  the  following  ex- 
amples. 

First,  let  it  be  proposed  to  extract  the  cube  root  of 
the  radical  quantity  26  +  15VT,  or  26  -f-  V675. 
Here,  A  being  =  26,   B  =  675,  and  72  =  3,   we  have 

..  _  26^s/67f\^  i  ^  —  _  3,732  ±  ,268 

''  2  2>c  26-f  V675T'  2  ) 

==  2  ;  and  tj  (  =  4  —  676  —  675")^)  =  3  ;  and  conse- 
quently  x  +  V  ?/  =  2  -f-  V  3  =  the  value  required  for 

-  +  V~3^  X  2  -f  v/3  X  T+v"3  =  26  -f.  \/675. 

Again,  let  it  be  required  to  extract  the  biquadratic 
root  of  161  -f-  V25920.      In  this  case,  A  being  =161, 

B  =  25920,  and  71  =  4,  we  have  x  (=  ^^^^^^^^^^r 

-  ^  -    l!!£ijL£i5)    ==    o,    and 


2    X    321,99,   Ssfc.]^ 


288  The  Application  of  Algebra 

t/  (=  4  +  25921  —  25920]*j  =  5  j  therefore  the  root 
sought  is,  here,  =  2  -f  V  5. 

Lastly,  if  it  were  required  to  find  the  first  sursolid 
root  of  76  +  V5808 ;    then,  by  proceeding  in  the  same 

manner,  x  will  be   found  (=    - — ^ ''- )  =  1,  and 

«/(=!—'  5776  —  5HOs]^)  =  3:  and  so  of  others. 
But  it  is  to  be  observed,  that  the  second  part  of  the 
Value  of  x^  to  which  both  the  signs  +  and  —  are  pre- 
fixed, is  to  be  taken  afilrmative  or  negative,  according 
as  that  or  this  shall  be  found  requisite  to  make  the  value 
of  X  come  out  a  whole,  or  rational  number :  and  that, 
if  neither  of  the  signs  give  such  a  value  of  x^  then  this 
method  is  of  no  use,  and  we  may  safely  conclude  that 
the  quantity  proposed  does  not  admit  of  such  a  root  as 
we  would  find.  It  may  also  be  proper  to  remark  here, 
that,  if  the  upper  sign  in  the  value  of  x  be  taken,  the 
tipper  sign  in  that  of  y  must  be  taken  accordingly  ;  and 
that  the  application  of  logarithms  will  be  of  use  to  facili- 
tate the  extraction  of  the  root  A  +  VB  f «,  as  being  suiH- 
ciently  exact  to  determine  whether  a:  be  a  whole  number, 
and,  if  so,  what  it  is. 

Thus  much  in  relation  to  the  extraction  of  the  roots 
af  radical  quantities  ;  it  remains  now  to  explain  the  man- 
ner of  taking  away  radical  quantities  out  of  the  deno- 
minator of  a  fraction,  and  transplanting  them  into  the  nu- 
merator. 

In  order  to  which,  supposing  r  to  denote  a  whole  num- 
ber, it  is  evident,  in  the  first  place,  that 

X''  —  y''  =~.ir^  X  ^^-1  +  x^'^y  +  x^^y^_  .  •  •  +  y'"'^ ; 
since,  by  an  actual  multiplication,    the  product  appears 

to  be  i  ^^  +  ^'""'^  +  ^"''^'  "^  ^'''^''  ^''        \    where  all 

the  terms,  except  the  first  and  last,  destroy  one  another. 
Hence,  by  equal  division,  we  have 


to  Geometrical  Problems.  289 

_J_^._r-+^^^+^-g^_+y-_     And,  in  the 

X  —  y  x^  —  y^ 

very  same  manner,  it  will  appear  that 

X  +y  x^ ±  y^ 

the  sign  —  or  +,  in  the  denominator,  takes  place,  accord- 
ing as  the  number  r  is  even  or  odd. 

Let  now  x  =  A***,  and  y  =  B^ ;    then  our 'equations 
will  become 

'j^^ni, Qn  J^rm  _  gj-ji  ^ 

J  Arm—tn  ___  ^rm— 2mJ^n    i    J^r}n—5mT^2n  ,  ,  ,  ^  -^  grn— ri 

From  which  theorems,  or  general  formulce^  the  mat- 
ter proposed  to  be  done  may  be  effected  with  great  far 

cility:    for,  supposing  —-——^,  or  ^,,  ^  ^^  to  be  a 

fraction  having  radical  quantities  A"",  B"  in  the  denomi- 
nator, it  is  plain,  that  its  equal  value,  given  by  the  said 
equations,  will  have  its  denominator  entirely  free  from  ra- 
dical quantities,  if  r  be  so  assumed  that  both  rwz  and  rn 
may  be  integers. 

To  exemplify  which,   let  the  fraction 


V2  —  1 

^=n be  propounded ;    then,  A  being  =  2,6  =  1, 

m  =  ^,  and  n  =  1,  we  shall,  by  taking  r  =  2,  have  (^from 
theorem  t)  r=-^^ =  ^  *'  '^^  =  VT+  1. 


2P 


290  The  Appiication  of  Algebra 

1 

Again,  let  the  given  fraction  be  --  _^.- 


or  : _^ ~7-      In  which  case.,  A  being  =  cx^ 

cx\-  +  <:"*  +  x""]^ 
B  1=  c^  +  jf'*,  w  =  |,  and  7i  =  1,  we  shall,  by  taking  r  =  4, 

have  =, 
c 


If  the  numerator  be  not  a  unit,  you  may  proceed 
in  the  same  manner,  and  multiply  afttrwards  by  the 
numerator  given.      Thus,  in  the  case  mentioned  at  the 

v  "2"-—  1 
beginning  of  this  scholium^  w^e  had  given ^ — ^^, 

2V  2  —  11 

1 


which  may  be  reduced  to  V  2  —   1    X 


-,  or  to  ^2  —  1    X 


2V  2   —  1 
1  1 


2V  2  —  1  Sl   —   1  8l  —  1 

2V^  -fl 


but is  found  (by  theorem  Vi  to  be  = — 

8|  —  1  ^  ^  ^  8—1 


2^  2  -4-  1       2  V  2  -f  1 


whence  our  expression  becomes  V  2  —  1 
:  which,  by  multiplication,  £5?c. 


7  7  _ 

5\/2~l 


is  reduced,  at  length,  to  ' 


49 


PROBLEM  XXVIL 

Having  one  leg^  AC,  of  a  right-angled  triangle  ABC, 
to  find  the  other  leg  BC,  so  that  the  hijpothenuse  AB  shall 
be  a  mean  proportional  between  the  perpendicular  Cli  fall- 
ing thereon^  arid  the  perimeter  of  the  triangle. 


to  Geometrical  Problems* 


291 


Put    AC  =  a^   and    BC  =  ;c ;    then    will    AB  = 
AC  X  BCT  ax 


Vxx  +  aa^  and  CD  (  = 


therefore,  by  the  ques* 
tion^  X  -{-  a  +  Vxx  -f  aa 
:  \^xx  +  aa',  :  Vxx  +  aa 

ax 
:  —  ;    and  conse- 

\\\x  +  aa 

ax^  4-  a^x 


quently 


+  ax  js^ 


V  XX  -J^  aa 

=    XX    -h    aa :      whence     a^x'^     +    2a^x^    +    a'^x^     =; 
XX  +  aa  —  axY  X  ^^  +  ^^*     Divide  by  a^x^  (according 

X  fl 

to    the    rule    at   page   156),    so    shall  —  +  2  -| = 

a  ^ 


X  a         '    P         X     ^     a 

-  +  --M   X  -  +  -z\ 


which,  by  making  z  = 


1 is  reduced  to  :s  +  2  =  z  —  l"|^  x  z,  or  z^ 

a  X 

—  22^  =  2.       This,    solved    (by  the    rule  for  cubics), 

gives  2;  =  £  +  -Lx  35+  v/TlGl  1^  +  —  X  4=r— 

^        ^  ^       35+\/116ir^ 


=  2,359304 :  whence  .v  (  =  —  X  2  ±  Vz^.  —  4)  will  like 
wise  be  known. 


PROBLEM  XXVIII. 

The  base  AB,  and  the  perpendicular  BE,  of  a  right- 
angled  triangle  ABE,  being  given  ;  it  is  proposed  to  find  a 
point  (C)  in  the  perpendicular^  from  whence  txvo  i'ight  lines 
C  A  and  CD  being  drawn^  at  right  angles  to  each  other ^^ 
the  tzvo  triangles  ACD  and  K^C^  formed  from  thence^ 
shall  be  equal. 

Suppose  DF  parallel  to  AC,  and  let  AF  be  drawn : 
putting  AB  =  a,  BE  ^b,  BC  =  v,  and  AC  {Va^  +  x''-) 


292 


The  Application  of  Algebra 


^  y.  Then,  since  FD  is  parallel  to  AC,  the  triangle 
ACF  will  be  equal  to  ACD,  or  ABC  ;  and  therefore 
CF  =  BC  =^5    whence  we  have  EF  (=£6  — BF) 

:=:b  —  2x,  and  EC(=EB 
—  BC)  =  ^  '—  •%• :  more- 
over, by  reason  of  the  si- 
milar triangles  ABC  and 
CDF,  we  have  y  (AC)  : 
.V  (BC)  :   :  X  (CF)   :   FD 

XX 

Whence,  because  of  the 
parallel  lines   AC  and  FD, 


(EF) 

hx^  — 


_^  it  wiU  be,  ~  (FD)  :  b  —  2x 

A  y  ^ 

y  (AC)   '.   b  —  X  (CE)  ;     and  consequently 


=  b  —  2x  X  y-,  or  bx^ 


■=.  b  —  2x  X  y^ 


which  equation,  by  writing  a^  +  a;^,  instead  of  its 
equal,  y^,  becomes  bx^  —  x^  =  ba^  +  bx^  —  ^a^'x  —  2x^ ; 
whence  we  have  x^  +  2a^x  =  ba^^  and  therefore  x  r= 

llb^~W~^  J^ — — . 


PROBLEM  XXIX. 

Three  lines^  AO,  BO,  CO,  drawn  from  the  angular 
points  of  a  triangle  to  the  center  of  the  inscribed  circle^ 
being  given  ;  to  find  the  radius  of  the  circle^  and  the  sides 
of  the  triangle. 

If  upon  CQ  produced,  the  perpendicular  BQ  be  let 
fall,  and  the  radii  OE,  OF  be  drawn  to  the  points  of 
contact,  the  triangles  BOQ  and  A  OF  will  appear  to  be 
equiangular ;  because  all  the  angles  of  the  triangle 
ABC  being  equal  to  two  right  ones,  the  sum  of  all  their 


to  Geometrical  Problems.  293 

halves,  OCB  +  OBC  +  OAF,  will  be  equal  to  one 
right  angle ;  but  the  two  former  of  these,  OCB  + 
OBC,  is  equal  to  the  external  angle  QOB  ;  therefore 
QOB  +  OAF  =  a  right  angle  =  QOB  +  OBQ, 
and  consequently 
OAF  =  OBQ. 
Putnow  AO=«, 
BO  =  ^,  CO  =  c, 
and  OF  =  xi 
then,  because  of 
the  similar  trian- 
gles, we  have  a  : 
X  :  :  b  :  OQ,  = 

— ;  whence  BQ2 

a 

bbxx 

■=z    b^    --^    , 

aa 

and  BC^  (=  CO^  +  BO^  +  2CO  x  OQ)  =z  b^  +  t" 

+  '^—.       But  BC2  :  BQ2  :  :  OC^  :   OE^ ;     that  is, 
a 

j:. [ — :  :   :  c^  :   x^.       l<rom 


whence  we  get  this  equation,  viz.  ax^  X  cib^  +  cic^  +  2bcx 

=  b^c^  X  d^  —  x^  \    which,  by  reduction,  will  become 

ab       ac      be         „      abc        ,  1      r        t 

x^  ^ f-  — 7  -f X  x^ziz  — :  wiience  a:  may  be  found, 

^  2c      2b      '^a  2  ^  ' 

and  from  thence  the  sides  of  the  triangle. 

If  two  of  the  given  lines,  as  OC  and  OB,  be  sup- 
posed equal,  the  result  will  be  more  simple :  for,  by 
writing  b  for  c  in  the  equation  ax^  X  cib^  +  ac^  4-  2bcx 
=  h^c^  X  a^  —  x^^  ^c.  we  shall  have  ab^x^  X  2a  +  2x 
=  ^^  X  d^  —  x"^  ;    which,  divided  by  ^^  ^  ^  _|,  ^,^  gives 

^Zax'^  ^  b^  X  a  ^ —  x\     whence  x^  -j =  \h^    and 

2a 


■4 


b^    .      ^^    _  ^  -  bv'Saa  4- bb  ^  bb 
1 6aa       4« "  4a 


294 


The  Application  oj  Algebra 


!  From  the  same  equations  the  problem  may  be  resolv- 
ed, when  the  distances  from  the  three  angular  points  to 
the  circumference  of  the  inscribed  circle  are  given :  for, 
denoting  the  said  distances  by  f^  g^  and  A,  you  will  have 
AO  z=z  X  +  f^  BO  =.  X  +g^  and  CO  z=.  x  +  h-,  which 
values  being  written  in  the  room  of  «,  b^  and  c,  there  will 
arise  an  equation  of  six  dimensions  :  by  means  whereof  x 
may  be  found. 

PROBLEM  XXX. 
To  draw  a  line  NM  to  touch  a  circle  D,  given  in  mag- 
nitiide  and  position^  so  that  the  part  thereof  KC^  intercepted 
by  txvo  other  lines  BK^  BL,  given  in  position^  shall  be  of  a 
given  iength. 

Suppose  CP  and  DE  to  be  perpendicular  to  AB,»and 
DF  and  DG  to  AC  and  PC,  respectively ;  and  let  DA, 
DC,  and  DP  be  drawn;    putting  DE  =  a,  DF  =  b^ 
PC  =  Xy  PA  =  ?/,  and  the  tan- 


gent of  the  given  angle  BCP  to  the  radius  1  =  t.  , 
Then,  by  trigonometry^  1  :  t  i  \  x  :  tx  =:  BV  ;  there-  ' 
fore  DG  (  =  ?E)  z=  d  —  tx -,     which,  multiplied  by 

IPC,  or  —,  gives  ^^  —  %^  for  the  area  of  the  tri- 
angle CDP:  in  like  manner,  the  area  of  the  triangle 
PDA  will  be  found  =  ^;  and  that  of  ADC  =  -|- ; 
which  three,  added  together,  are  equal  to  the  whole  area  ^ 


ACP  ',   that  is, 


dx  —  tx^ 


ay 


be 


+  —  +  — = 


_  ^y. 


and  conse- 


to  Geometrical  Problems,  2d5 

queiitly  be  +  dx  —  tx^  =  xy  —  ay.  Let  both  sides 
of  this  equation  be  squared,  and  you  will  have 
'  be  4-  dx  —  tx^'\  :=  X  —  tt  I  X  y^  -=  oc  —  0]^  X 
c^  —  x^  ;  that  is,  b^c^  +  2bedx  —  2betx^  +  d^x^  —  2dtx^ 
+  ^2^.4  =  — ^  +  2ax_^__—jn^^  +  e^x^  —  2ac^x  4-  a\^  ; 
whence  1  -¥■  c^  X  x"^  —  2a  +  2at  X  ^^  +  a^  —  c^  -f  (^^  —  ^^^^ 
X  ^^  +  2ac2  +  2/bcY/  X  X  —  6zV  ^  ^2^2  __  Q  .  fj.Qj^  which 
the  value  of  x  may  be  found;  and  then,  the  value  of  y 
(  =  Vc^  —  jv^)  being  known,  the  position  of  the  points  A 
and  C,  through  which  the  line  must  pass,  will  also  become 
known. 

If  the  given  angle  B  be  a  right  one,  the  point  B  will  co- 
incide with  P;  and,  therefore,  t  in  this  case  beinp^  =  0, 
the  equation  will  become  x^  —  2ax^  +  a^  —  c^  -j-  d^  X  x^ 
-L  2^6-2  +  2bed  x  x  —  a^c^  +  b^e^  =  0. 

When  the  circle  touches  the  right  line  AB,  a  will 
then  be  equal  to  b :  and,  in  that  case,  the  equation 
will  be  1  -f-  ^^  X  x^  —  2a  -{- 2di  X  x^  +  a^  —  c^  -f-  ^^  —  2act 
X  X  +  2ae^  +  2aed  =  0,  because  the  two  last  terms  —  a^c^ 
+  b^e^  destroying  each  other,  the  whole  may,  here,  hp  di- 
vided by  a;. 

Lastly,  if  ^  be  =  0,  or  the  line  AC,  instead  of  touchlag 
a  circle,  be  required  to  pass  through  a  given  pouit,  the 
equation  will  then  become  1  -^  t^  x  x^  —  2a  +  2dtx  x^ 
+  a^  —  c^  -{-d^x  x^  +  2ac^x  —  a^c^  =  0. 

PROBLEM  XXXL 

Supposing-  AGi perpendicular  to  AF,  and  the  given  rig/it 
line  AF  (50)  to  be  divided  into  Jive  equal  parts^  in  the 
points  B,  C,  D.andE  ;  to  find  a  point  P  in  the  perpendi- 
cular AGi^from  which^  ifj^'^e  right  lines  be  drawn  to  the 
points  B,  C,  D,  E,  and  F,  the  sum  of  the  outermost  PF  + 
PE  shall  be  equal  to  the  sum  of  the  three  innermost  PD  4^ 
PC  +  PB.  ^ 

Put  AP=:;y;  then  {by  Euc.  47.  1.)  BP  rrr  yioo  -f  x"^ 
^^  =  ^^400  +  x^^  csPc.  and,  consequently,  VlOO  +  x^ 

+  V400  +  x2   +  V900  ^  x^  ^~  vTeoo"  4-  x"^  ~ 


2:96  The  Application  of  Algebra 


V2500  +  ^^  =  0.  Now,  by  reflecting  a  little  on 
the  nature  of  the  problem,  it  is  easy  to,  perceive  that 

IQ  PF4-PE  must  be 
Ip  greater  than  90,  see- 
^^  ing  AF  +  AE  is  = 
90 ;  whence  it  ap- 
pears that  PD  -f. 
PC  +  PB  must  also 
exceed  90,  and  that 
PC  (considered  as  a 
mean  between  PD 

-^ r^ p        ^       ^        and  PB)   must  be 

^      ^       ^        ^  greater    than    30; 

hence  I  conclude,  that  the  value  of  AP,  as  it  is  some- 
thing less  than  PC,  will  be  somewhere  about  30; 
and  therefore  I  write  30  +  e  for  x;  and  then,  rejecting 
all  the  powers  of  e^  above  the  first,  as  inconsiderable, 
our  eqaation  stands  thus,  VlOOO  +  60^  -f  Vl300  +  60^. 
+  V1800  +  eOe-  —  V2500  -f  60^  —  V3400  +  60^  =  0  ; 
which,  by  the  method  explained  in  page   174,  will  be 

transformed  to  vToOO    +  ^^^^^  X   ^^     .     V1300    + 
^     100  ^ 

^lli2L>Ll^  +  vr8o5  +  )^im.2<Jl  _  50  -  6.  ~ 

130  180 

N/ 3400  X  3^ 

V3400  — =  0  ;  this,  contracted,  gives  1,8  + 

1,37^  =  O;  whence  ^  =  —  1,3,  and  consequently  x  = 
28,7,  nearly.  Let,  now,  28,7  be  put  =  x  ;  and  then,  by 
proceeding  as  above,  we  shall  have  ,0083  +  1,43^  =  O; 
hence  ^  =  —  ,0058,  and  x  =  28,6942 ;  which  is  true  to 
the  last"  ii^ure. 

PROBLEM  XXXIL 

The  perimeter  AB  ~f-  BC  +  AC,  and  the  perpendicular 
CP,  of  a  triangle  ABC,  whose  sides  are  in  harmonic  pro- 
portion (AB  :  BC  :  :  AB  —  AC  :  AC  ~  BC)  being  giv- 
en ;  to  deter  mint  the  triangle* 


to  Geometrical  Problems. 


297 


Let  abc  be  another  triangle,  similar  to  the  proposed 
one :  and  let  ab  =  1,  be  :=.  x^  ac  -=1  y,  CP  =  a,  and 
AB  +.BD  +  AC  =  b :  then,  half  the  sum  of  the  three 

2  ' 

same  each  particular  side  be  subtracted,  and  all  the  re- 


sides of  the  triangle  abc  being 


if  from  the 


mainders    be    multiplied    continually  together,    and  that 
product,    again,    by  the    said  half   sum,   we  shall  have 

1  -h^  —  y^  1  +^  — ^  w?/  +  ^— i^l±iL±^-,equal 


2  '  2  2  2 

to  the  second  power  of  the  area  abc  {by  prob.  15)  : 
which,  as  the  base  is  unity,  also  expresses  \  of  the  square 
of  the  perpendicular.  But  the  squares  of  the  sides,  as 
wellas'  the  sides  themselves  of  similar  triangles,  are  pro- 
portional, and  therefore  1  -f  c^:  4-  y  r  •  ^^  •  ^ 
1  -f-^  —  y  X  1  —  X  +  y  X  y  -^-x  —  1  Xl  +x  +  y  ^    ^ 


whence  we  have  4a^  x  l+:v  +  z/=l  +x  — -^/Xl' —  ^  +  y 
Xy  -yx  -^1  X  bb\  but  the  sides  AB,  AC,  and  BC,  be- 
ing given  in  harmonic  proportion,  therefore,  1 ,  y^  and  x^ 
must  likewise  be  in  the  same  proportion ;  that  is,  1  :  x 
:  :  1  —  y  -  y  — '  ^  ;  whence  y  —  x  =^  x  —  xy^  and  there- 

fore   y    =    ;      which,    substituted    above,    gives 


1    -^  X- 


Aa^Xl  -{-4x4.  x^ 
1  -f  ^        "~ 


lJ-^3        1+2a?— AT^        2x  +  x^ 1 

X  ■    \    . X         ' — — 


1   +  X 


1  +x 


\  -^x 


X  ^%or4a^x  1  4-4;^  +  ;^^  X  1  +  ^T  =  1  +  ^''X  1  +  2;\: — x^ 
X  2Ar  4-  a:*  —  1  X  6^ ;  from  which  x  will  be  found,  and 

2Q 


298 


The  Application  of  Algebra 


2x 


also  y  (  = •) ;   and  from  thence  the  reqAired  sides 

of  the  similar  figure  ABC  will,  by  proportion,  be  likewise 
known. 


PROBLEM  XXXIIL 

Let  there  be  three  equi-different  arches^  AB,  AC,  and 
AD  ;  and^  supposing  the  sine  and  co-sine  of  the  mean  AC, 
of  the  lesser  extreme  AB,  and  of  the  common  difference  BC 
{or  CD)  to  be  gvuen^  it  is  proposed  to  find  the  sine  and  co- 
sine of  the  greater  extreme  AD. 

Upon  the  radius  AO  let  fall  the  perpendiculars  B^, 
Cc,  and  D^;  join  B,  D,  and  from  the  center  O  let 
the  radius  OC  be  drawn,  cutting  BD  in  n-^  also  draw 
nR  parallel  to  Cc, 
meeting  AO  in  R : 
then,  because  of  the 
similar  triangles  OCc 
and  OwR,  it  will  be, 
OC:On:  :  Cc:nR; 
and  OC  :  On  :  :  Oc  : 
OR :  whence  we  have 


^R  = 


Ccx  On 


,  and  OR  = 


Ocx  On 


O 

but,  since  BC  is 


OC     '  OC 

equal  to  CD  (and  therefore  B^z  equal  to  D^z),  nR  will, 
it  is  plain,  be  an  arithmetical  mean  between  B^  and  D<^, 


and  so  is  equal  to  half  their  sum,  or 


Bb  +  T^d 


and,  for 


the  very  same  reason,  OR  will  be  equal  to 
B^  +  T)d 


Ob  +Od 


consequently, 
Oc  X  On 


whence  Dd  = 


Cc  X  On        ,  Ob  +  Od 
-OC-'  and  —^ 

Cc  X  20n 


OC 


-^  Bb,  and  OD  = 


Ob ;  which,  if  the  radius  OC  be  supposed 


OC       ' 
Oc  X  ^On 
OC 

unity,  will  become  Dd  =  Cc  x  20n  —  B^ ;  and  Od  = 
Oc  X  20n  •—  O^ :  from  whence  we  have  the  two  follow- 
ing theorems. 


to  Geometrical  Problems,  299 

Theor.  1.  If  the  sine  of  the  ?nean  of  any  three  equi'dif 
ferent  arches  (the  radius  being  supposed  unity)  be  multi- 
plied by  twice  the  co-sine  of  the  common  difference^  and  from 
the  product  the  sifie  of  either  extreme  be  subtracted^  the 
remainder  will  be  the  sine  of  the  other  extreme. 

Theor.  2.  And  if  the  cosine  of  the  7nean  of  three  equi- 
different  arches  be  multiplied  by  twice  the  cosine  of  the  com- 
mon difference^  and  the  cosine  of  either  extreme  be  sub- 
tracted from  the  product^  the  rernainder  will  be  the  cosine 
of  the  other  extreme^ 

PROBLEM  XXXIV. 

The  sine  and  cosine  of  an  arch  being  given^  to  find  the 
sine  and  cosine  of  any  multiple  of  that  arch. 

Let  the  given  arch  be  represented  by  A,-  its  sine  by  x^ 
and  co-sine  by  |y,  the  radius  being  unity.  Then,  since 
the  arch  A  may  be  considered  as  an  arithmetical  mean 
between  0  and  2A,  we  shall,  by  the  first  of  the  two  pre- 
ceding theorems,  have 

sine  of  2  A  (=  sine  of  A  X  ^  —  sine  of  O)  =  xy  ; 
sine  of  3  A  (=  sine  of  2  A  X  ^  -^  sine  of  A)  =  xy^  •—  x  ; 
sine  of  4A  (=  sine  of  3 A  X  ?/  —  sine  of  2 A  —  xy'^  —  xy 

—  xy)  =  xy^  —  2xy  ; 
sine  of  5  A  (=  sine  of  4 A  X  y  —  sine  of  3  A  =  x?j*  —  2xy' 

'— '  xy^  +x)  =  xy^  —  3xy^  +  x  ; 
sine  of  6 A  (=  sine  of  5  A  x  y  —  sine  of  4 A  =  xy^  —  3iXy^ 

+  xy  ^-^  xy^  -f-  ?>xy)  =  xy^  —  4xy^  +  2xy  ; 
sine  of  7 A  (=  sine  of  6A  x  y  —  sine  of  5 A  =  xy^  —  4xy^ 

+  3xy^  -^  xy^  -f  2xy^  — .  :v)  =  xy^  — •  5xy^  +  6xy^  —  x  : 
whence,  universally ^  the  sine  of  the  multiple  arch  72 A, 
where  n  denotes  anj^  whole  positive  number  whatever, 
will  be  truly  expressed  by  ;c  x  into  this  series : 

„  -    n— 2      ^»  .  72-^3    n — 4      ^  .  n — 4    n — 5    n — 6    "     ^ 
2^-1 — ^Xy^'+'-^x-Y-xy—^^'^r^'^T'^^ 
&?c.       Moreover,    from  the  second  theorem,    we  have 
co-sine  of  2 A  (c=  co-sine  of  A  x  y  —  co-sine  of  0  = 

y'^t^^y"  —  ^. 


r2 


SOO  The  Application  of  Algebra 

Co-sine  of  3  A  (  =  co-sine  of  2 A  X  y  —  co-sine  of  A  = 

y^  ^.j—Ia^  y'  —  ^y . 

T       ^        2^""^       2        ' 

Co-sine  of  4a  (  =  co-sine  of  3  A  X  y  -^  co-sine  of  2  A  = 

2  "~"2      '^■^  2 

whence,    universally^    the    co-sine  of  the  multiple  arch 

^^A    will    be    truly    represented    by  —  •—  -2 —   -j 

•^  >i  ^ 

7Z' —  3         ^.  n       n  —  4      w  —  5        ^  ^       n 

X  —^xy--^  —  -x—^X—^  xy^'^  +  -X 

X X  - — —  X  v"""^,  ^c.  which  series,  as  well 

2  3  4 

as  that  for  the  sine,  is  to  be  continued  till  the  indices  of  y 

become  nothing,  or  negative. 

But,  if  you  would  have  the  sine  expressed  in  terms 

of  X  only^  then,  because  the  square  of  the  sine  +  the 

square  of  the  co-sine  is  always  equal  to  the  square  of  the 

radius,  and,  therefore,  in  this  case,  x^  +  %y^  =  1,  it  is 

manifest  that  the  sines  of  all  the  odd  multiples  of  the 

given  arch  A,  wherein  only  the  even  powers  of  y  enter, 

may  be  exhibited    in   terms  of   x    only,    without    surd 

quantities :     so  that  4  —  4x^  being  substituted  for  its 

equal  2/^,  in  the  sines  of  the  aforementioned  arches,  we 

shall  have 

1st.  Sine  of  3 A  =  3a:  —      4x^ ; 

2d.   Sine  of  5A=:  5x  —    20^2  +    16x^ ; 

3d.   Sine  of  7A=z7x  —    56x^  +  112;^^  —  64^^  ; 

4th.  Sine  of  9A  =  9a?  —  120^:^  +432^^^  —  576x^  +256x^; 

And,  generally^  if  the  multiple  arch  be  denoted  by 
7iA,  then  the  sine  thereof  will  be  truly  represented  by 
n       n^  —  1  n       r?  —  1      n^  —  9        . 

n       n^  —  1      n^  —  9      n^  —  25  ^^    7   ,   ^       n^  ~  1 

7i2  _  9      71^^25      n*  —  49  ^      ... 
,     ^     X     ^    ^     X  ■    ^    Q     X  ^S  ^c. 
4.5  6.7  8.9 


to  Geometrical  Problems.  301 

•From  this  series,  the  sine  of  the  sub-multiple  of  any 

arch,  where  the  number  of  parts  is  odd,  may  also  be 

found,  supposing  {s)  the  sine  of  the  whole  arch  to  be 

give;i :  for  let  x  be  the  required  sine  of  the  sub-multiple, 

and  71  the  number  of  equal  parts  into  which  the  whole 

arch  is  divided;  then,  by  what  has  been  already  shown, 

,    „  _  n       11^  —  1         o   .    w       n^  —  1 

we  shall  have  nx X  -r — --  X  ^^  +  ,    X  X 

1  2.3  1  2.3 

X  ^^,  ^c,  =  5  :  from  the  solution  of  which  equa- 

tion  the  value  of  x  will  be  known.  Hence,  also,  we 
have  an  equation  for  finding  the  side  of  a  regular  po- 
lygon inscribed  in  a  circle  :  for,  seeing  the  sine  of  any 
arch  is  equal  to  half  the  chord  of  double  that  arch,  let 
^v  and  Iw  be  written  above  for  x  and  ^,  respectively,  and 

TIV  JZ  71^  ,        1  ^3 

then  our  equation  will  become  —. ■  —  X  — X  — 

^  2  12.38 


71 

77.2  - 

-1 

92  ► 

—  9 

v' 

,&fc.= 

71 

+ 

—  X 

— - 

X 

X 

T^l 

:|t^,  < 

or  Tiv 

X 

1 

2  . 

3 

4 

.5 

32 

1 

V^' 

—  1 

v^ 

n 

7?,^- 

-1 

^^^  — 9 

v' 

X  — 

+ 

X 

X 

X  —: 

,&fc.: 

=  ru; 

ex- 

2 

.  3 

4 

1 

2  . 

3 

4.  5 

16' 

pressing  the  relation  of  chords,  whose  coiTesponding 
arches  are  in  the  ratio  of  1  to  n.  But,  when  the  greater 
of  the  two  arches  becomes  equal  to  the  whole  peri- 
phery, its  chord  (w)  will  be  nothing,  and  then  the  equa- 
tion, by  dividing  the  whole  by  nv,  yfill  be  reduced  to 

2.3^4'*"2.3^4.5^16~  "TTT     ^ 

n2  — 9      72^  —  25       Z)«     ^.         ^        ,  .      , 

X  rr,  ere.  =  0 ;  where  72  is  the  num- 


4.5  5.6  64 

ber  of  sides,  and  v  the  side  of  the  polygon. 

From  the  foregoing  series,  that  given  by  sir  Isaac 
NewtOHy  in  Ph'iL  Trans,  mentioned  in  p.  242  of  this  Trea- 
tise, may  also  be  easily  derived.  For,  if  the  arch  A  and 
its  sine  x  be  taken  indefinitely  small,  they  will  be  to 
one  another  in  the  ratio  of  equality,  indefinitely  near,  by 


.302  The  Application  of  Algebra 

what  has  been  proved  at  p.  246 ;  in  which  case,  the  ge- 
neral expression,  by  writing  A  instead  of  x^  will  become 

uK X X  A^  +  —  X X 

1  2.3  1  2.34.5 

Therefore,  if  tz  be  now  supposed  indefinitely  great,  so 

that  the  njultiple  arch  nK.  may  be  equal  to  any  given 

arch  2,  the  squares  of  the  odd  numbers,  1,  3,  5,  £sPc.  in 

the  factors  ?i^  —  1,  n^  —  9,  r^  —  25,  &Pc.  may  be  rejected 

as  nothing,  or  inconsiderable,  in  respect  of  r^ ;  and  then 

72^  A'  n^A^ 

the  foreoxjinc;  series  will  become  tzA  — 1 

^      ^  2.  3^2.3.4.5 

,  £s?c.  wherein,  if  for  nA,  its  equal, 


2.3.4.5.6.7 
e,  be  substituted,    we    shall    then  have  %  - — 


2.3 

,  l^c.  which  is  the  sine 


2.3.4.5         2.3.4.5.6.7 

of  the  arch  z,  and  the  same  with  that  before  given. 

Moreover,  the  aforegoing  general  expressions  may  be 

applied,  with  advantage,  in  the  solution  of   cubic,  and 

certain  other    higher  equations,    included  -in  this  form, 

n  —  3         o«4         n  — 4      n  —  5 

VIZ.  2"  —  02"-*  H X  c^z"^-^ X  X 

'In  2n  Sn 


For,  if  z  be  put  =  y  \}' — ,  the  equation  will  be  transform- 

— I— 
cd  to  —       X  into  this  series 

71   I 

»      71      71' — 3  ^      71      n--^^     n*-^S        „__^ 

yn^ny^-^+j  X -^  X  y'^'  +  j  X  -^  X  -^  X  ^"^ 

TyTi  ^  7t  7t 

^c.  =/,  and  consequently  ^  —  —  X  y^"^  H X  - 


2  2        '^  2  2 


to  Gaometrical  Problems.  203 

X  y""^—  ~  X  ^^-^-^  X  ^■=^X^/'^-^£sPc.===^l4"  -.  from 

n  I 
whence,  as  it  is  proved  above,  that  the  former  part  of  the 
equation  (and  therefore  its  equal)  represents  the  co-sine  of 
n  times  the  arch  whose  co-sine  is  |z/,  we  have  the  follow- 
ing rule : 

Find^from  the  tables^  the  arch  whose  natural  cosine  is 
JL  jf  ,  71  a 

~=r-  {or  its  log",  co-sine  =  log*  if*"^  —  ^og:  — )  the  radius 
fjj  2  n 

n\ 
being  unity  ;  take  the  72th  part  of  that  arch^  and  Jlnd  its 

cO'siney  which  multiply  by  2y — ,  and  the  product  xvill  be 


n 


-2 


the  true  value  of  z^  in  the  proposed  equation  z"  —  az^"'- 

A X  d^z^-^ .—  X    X  a^z"^-^^  ks?c\ 

2n  2n  2n 

Thus,  let  it  be  required  to  find  the  value  of  z,  in  the 
cubic  equation  2^  —  4322  =  1 728 ;  then,  we  shall 
have  n  =  3,  a  =  432,  and  f  =   1728;     consequently 

.  ~J^-  (=  )  =  ,5,    and   the    arch    corresponding 

aj^  144jir 

71   I 

thereto  ==  60° ;    whence  the  co-sine  of  (20°)  ^  thereof 
will  be  found   ,9396926;     and    this,    multiplied    by  24 


=  2\/^)'Si 


(=  2\} — ),  gives  22,55262  for  one  value  of  z.      But 
^  n 

besides  this,    the    equation    has   two    other   roots,   both 

of  which  may  be  found  after  the  very  same  manner; 

for,  since  0,5  is  not  only  the  co-sine  of  60°,  but  also  of 

60°  +  360°,  and  60°  -f-  2  x  360°,  let  the  co-sine  of  (140°) 

^  of  the  former  of  these  arches  be  now  taken;    which 

is  — •  ,7660444,  and  must  be  expressed  with  a  negative 

sign,  because  the  arch  corresponding  is  greater  than  one 

right    angle,   and    less    than    three,       Then^   the  value 


304  The  Application  of  Algebra 

thus  found  being,  in  like  manner,  multiplied  by  24 
(  =  2y — ),  we  shall  thence  get  —  18,38506  for  ano- 
ther of  the  roots ;  whence  the  third,  or  remaining  root 
will  also  be  known ;  for,  seeing  the  equation  wants  the 
second  term,  the  positive  and  negative  roots  do  here  mu- 
tually destroy  each  other;  and  therefore  the  remaining 
root  must  be  —  4,16756,  the  difference  of  the  two  for- 
mer, with  a  negative  sign. 


PROBLEM  XXXV. 

From  a  given  circle  ABCH  it  is  proposed  to  cut  off  a 
segment  ABC,  such^  that  aright  line  DE  drawn  from  the 
middle  of  the  chord ^  AC,  to  make  a  given  angle  therewith^ 
shall  divide  the  arch  BC  of  the  semi-segment  into  two  equal 
farts  BE  and^C. 

Let  the  chord  BC  be  drawn,  and  upon  the  diameter 
HDB  let  fall  the  perpendicular  EF :  put  the  radius  OB 

of  the  circle  =  r,  and  the 
tangent  of  the  given  angle 
CDE  (answering  to  that 
radius)  =  f,  and  let  OF 
=  2  ;  then  will  EF  = 
V  rr  -^2,  and  BC  (  = 
2EF)  =  2V/rr  —  22,  and 
consequently  BD  (  = 
BC2  .  4r^  —  42^ 


BH^ 

^  .  r  -j-z 


2r 


from 


r-f-22X  r- 


which  taking  BF  =  r  —  2,  we  have  DF  = 

But,  by  trigonometry,  EF  :  DF  :  :  rad.  :  tang.  DEF, 

that  is,  v/TT^ii  :  ^-f  ^^'^  — ^  :  :  r  :  t.     Whence 


we  have  r  +  2zY  X  r  ~  2?  =  /^  X  r  —  z^ ;   where 


td  Geometrical  Problems.  305 

the  whole  being  divided  by  r  —  z,  there  results  r  +  22"]^ 
X  r*— 2  =  ^^  X  r  +  z:  which,  ordered,  gives  z^  — • 
^rr  —  tt  r 


X  z^-^  X  rr  —  tt. 


4  4 

Zrr  —  tt 


Put =  «,  and —  X   rr  —  tt  =  fi    then  it 

4  4  *^ 

will  be  2^  —  az  =/.  Therefore  find,  from  the  tables, 
the    arch    whose    co-sine    is    — ^ —    (the  radius  being 

unity)  ;  take  f  thereof,  and  find  its  co-sine  ;  which,  mul- 
tiplied by  2V^a^  gives  the  true  value  of  z,  (See  the  last 
problem.) 

Thus,  for  example,  let  the  radius  OB  (=  r)  =  1, 
and  the  given  angle  CDE  =  25°,  whose  tangent  (?)  is 
therefore  =  ,4663  ;  whence  ^a  =  ,23188,  and  if  = 
,09782. 

Now,  by  logarithms^  it  will  be  log,  if  — ■■  log.  ^a  — • 

|log.   ia  =    —   1.9425328    =    log.    — ~=i    =  log.   co- 

sine  of  28°  50' ;  whereof  the  third  part  is  9°  36f ',  whose 
log.  co-sine  (to  the  radius  1)  is  ^ —  1.9938609;  which 
added  to  the  ilog.  of  ^a  ( =  —  1.6826316)  gives 
—  1.6764925  =  log.  of  0,47478,  whose  double  ,94956, 
is  the  true  value  of  2,  or  FO  :  whence  the  correspond- 
ing arch  BE  =  18°  16^',  and  consequently  BC  (=  2BE) 
=  36°  33'.  By  means  of  this  problem,  that  portion  of  a 
spherical  surface  representing  the  apparent  figure  of  the 
sky  is  determined. 

PROBLEM  XXXVI. 

The  base  AB,  and  the  difference  of  the  angles  at  the  baic 
being  given^  while  the  angles  themselves  vary;  to  find  the 
locus  of  the  vertex  E  of  the  triangle. 

Let  the  base  AB  be  bisected  in  O,  and  the  angle  BOD 
so  constituted  as  to  exceed  its  supplement  AOD  by  the 
given  difference  of  EAB  and  EBA;  and  let  ED,  APQ, 
BSF  be  perpendicular,  and  EPF  parallel  to  OD  : 
then,    since  the  angle    BCE    (BOD)  as  much  exceeds 

2R 


S06 


The  Application  of  Algebra 


ACE,  as  CAE  exceeds  CBE,  it  is  evident  that  the  sum 
of  the  two  angles   BCE,   CBE,  of  the  triangle   BCE, 

is  equal  to  the  sum 
of  the  two  angles 
ACE,  CAE  of  the 
triangle  ACE  ;  and 
consequently,  that 
the  remaining  an- 
gles AEC  and  BE  C, 
are  equal  the  one  to 
the  other :  there- 
fore, by  reason  of 
the  similar  triangles 
EFB,  EAP,  we 
have  EF  :  EP  :  : 
BF  :  AP,  that  is, 
OD  +  OQ  :  OD 
—  OQ  :  :  QA  + 
DE  :  QA  —  DE  ; 
whence,  by  com- 
position and  divi- 
sion, 20D  :  20Q 
•  :  2QA  :  2DE; 
wherefore  OD  X  DE  is  =  OQ  X  QA  ;  which  is  the 
known  property  of  an  equilateral  hyberbola  with  respect 
to  its  asymptote. 


PROBLEM  XXXVII. 


To  Jind  the  solidity  of  a  conical  iingula  BFCB,  cut  off 
hy  a  plane  BRFSB  passing  through  one  extremity  of  the 
base  diameter. 

Let  EPF  be  parallel  to  the  base  diameter  BC,  cut- 
ting AD  the  axis  of  the  cone  in  P  ;  also,  let  An  be  per- 
pendicular to  BF  ;  join  P,  72,  and  let  RS  be  the  conju- 
gate axis  of  the  elliptical  section  BRFSB :  then  the 
part  ABF,  above  the  said  section,  being  an  oblique  el- 
liptical cone,  its  solidity  will  be  expressed  by  ,7854  X 

SR  X  BF  X  — ,  that  is,  by  the  area  of  its  base  BRFSB 


to  Geometrical  Problems* 


507 


drawn  into  one-third  of  the  perpendicular  height.  But 
the  triangles  BCF  and  AP/z  will  appear  to  be  equi-an- 
gular ;  for,  APF  and  A?zF  being  both  right  angles, 
the  circumference  of  a  circle,  described  on  the  diame- 
ter AF,  will  pass  through  P  and  w;  and  so  the  angles 
AF?z  (BFC)  and  AP;2,  as  well  as  AFP  (FCB)  and  AnV, 


insisting  on  the  same  arcli,  are,  respectively,  equal. 
Hence  we  have  BC:BF::A/z:AP;  and  therefore 
BF  X  A?2  =  BC  X  AP:  this  value  being  substituted  above, 
the  content  of  the  part  ABF  becomes  SR  x  BC  x  AP  x 
,2618  :  which,  because  SR  is  known  to  be  =  VBC  x  l^F, 
is  farther  reduced  to  BC  X  AP  X  VBC  x  KF  X  ,2618. 
This,  subtracted  from  BC^  X  AD  x  ,2618,  the  content 
of  the  whole  cone  ABC,  leaves 

BC^  X  AD  —  BC  X  aFx  VBC  x  H  X  ,2618  for  the 

required  solidity  of  the  singula  BCF  ;    which,  because 

DP  X  BC        ,    .p        DP  X  EF       .,,  , 

r,  and  AP  =  -^-^^ _-_,  will  be  re- 


AD  = 


BC  — EF' 


BC  — EF' 


duced  to 


BC^  — EFx  VBCxEF 
BC  —  EF 


X  ,2618  DP  X  BC. 


508 


The  Application  of  Algebra 


PROBLEM  XXXVIII. 

Let  A  and  B  be  two  equal  weights^  made  fast  to  the  ends 
of  a  thready  or  perfectly  fiexihle  line  pVnGiq^  supported  by 
two  pins  ^  or  tacks^  P,  Q,  in  the  same  horizontal  plane ;  over 
which  pins  the  line  can  freely  slide  either  way  ;  and  let  C 
be  another  weight^^  fastened  to  the  threhd^  in  the  middle^  be- 
tween P  and  Q :  nozu  the  question  is^  to  find  the  position  of 
the  weight  C,  or  its  distance  below  the  horizontal  line  PQ, 
to  retain  the  other  ttvo  weights  A  and  B  in  equilibrio. 

Let  PR  (  =  iPQ)  be  denoted  by  a,   and  Rn    (the 
distance  sought)  by  x ;  and  then  Pn,  or  Qn,  will  be  re- 


presented by  Vrt^  +  x^.  Therefore,  by  the  resolution  of 
forces^  it  will  be,  as  V  a^  -f  x^  (P?z)  :  x  (R/z)  :  :  the 
whole    force    of   the    weight   A  in  the  direction   Pn,  : 

A.X 
"   '  •    "     -,  its  force  in  the  direction  tzR,  whereby  it  en- 
Va^  +  x^ 

deavours  to  raise  the  weight  C  ;  which  quantity  also  ex- 
presses the  force  of  the  weight  B  in  the  same  direction  : 
but  the  sum  of  these  two  forces,  since  the  weights  are  sup- 
posed to  rest  in  equilibrio,  must  be  equal  to  that  of  the 

2Av 
weight  C  ;  that  is,  —  =  C  ;  wheiice  w^e  have  4A^x^ 


:=:  C^fl^  +  C^;^^,  and  consequently  x  \ 


aZ 


V4A2. 


to  Geometrical  Problems^ 


309 


PROBLEM  XXXIX. 

To  determine  the  position  of  an  inclined  plane  AE,  along 
which  a  heavy  body^  descending  by  the  force  of  its  own  gra- 
vity from  a  given  point  A,  shall  reach  a  right  line  BP, 
given  by  position^  in  the  least  time  possible* 

Through  the  given  point  A,  perpendicular  to  the 
horizon,  let  there  be  drawn  the  right  line  RB,  meeting 
BP  in  B ;  also  conceive  the 
semicircle  AER  to  be  de- 
scribed, touching  BP  in  E  ; 
then  let  AE  be  drawn,  which 
will  be  the  position  required ; 
because  the  time  of  descent  a- 
long  the  chord  AE  being  equal 
to  that  along  any  other  chord 
A/z,  it  will  consequently  be 
less  than  the  time  of  the  descent 
along  A^,  whereof  An  is  only 
a  part:  therefore,  if  AQ  and 
OE  be  now  made  perpendi- 
cular to  BP,  we  shall  have  (by 
reason  of  the  similar  triangles) 
AB  :  AQ  :  :  AB  +  AO  : 
(OE)  AO  ;  whence,  by  multiplying  extremes  and  means, 
AB  X  AO  =  AQ  X  AB  +  AQ  x  AO  ;  therefore  AB 
X  AO  —  AQ  X  AO  =  AQ  X  AB,  and  AO  (OE)  = 

-^--r -r-r^ ;  from  which  BE  and  AE  are  also  given. 

xV.JlS  —  AQ 

The  geometrical  construction  of  this  problem  is  ex- 
tremely easy  ;  for,  if  AQ  (as  above)  be  drawn  perpen- 
dicular to  BP,  and  the  angle  O AQ  be  bisected  by  AE, 
the  thing  is  done  :  because,  OE  being  drawn  parallel  to 
AQ,  the  angle  OE  A  is  =  QAE  =  EAO  ;  and  so,  AO 
being  =  OE,  the  semicircle  that  touches  BP  will  pass 
through  A. 


310  llie  Application  of  Algebra 


PROBLEM  XL. 

A  ray  of  lights  from  a  lucid  point  P,  hi  the  axis  AP,  of 
a  concave  spherical  surface^  is  reflected  at  a  given  point  E 
tn  that  surface  ;  tofnd  the  point  D  w  here  the  refected  ran 
Tneets  the  axis. 

Draw  EQ  perpendicular  to  AP,  and  from  the  center 
C  let  CE  be  drawn;    also  make  CE  r=  a^  CQ  =  b^  CP 

=     c,    CD    =    X  ; 
-R  \mA(byEucA2.2>.) 

PE  will  be  r= 
VV  +  c2  +  2bci 
wherefore,  the  an- 
gles   of   incidence 

andreflection  CEP 

D       Q      A     and    CED    being 
equal,  we  have,  as 

PC   (c  :   CD   {x)  :   :   PE   (Va^  +  c?  +  ^Z^^)  :  ED  = 

cc\/  a^   -^  c^  -^  2bc         1  r        .1 

— ;     also,    for    the  same    reason,    wc 

have    PE    X    ED    —    PC    X    CD    =    EC%    that    Is, 

--2 ■ ■ ex  =  a^  ;    which,  reduced,   gives 

ca^ 
X  =  ^ -—    showino;  how  far  from  the  center  the  rav 

cuts  the  axis.  But  if  the  lucid  point  P  be  supposed  in- 
finitely remote,  so  that  the  ray  PE  may  be  considered  as 
parallel  to  the  axis  AP,  the  expression  will  be  more  sim- 
ple ;  for  then  (2^,  in  the  divisor,  may  be  rejected  as  nothing 
in  comparison  of  23c ;    that  being  done,   CD,  or  x^  be- 

comes  =  — - :    which,  therefore,  if  E  be  taken  near  the 
2b  '  ' 

vertex  A,  will  be  =  -ia,  very  nearly. 

PROBLEM  XLL 

To  fnd  the  magnitude  and  position  of  an  image  formed 
hij  refraction  at  a  given  lens. 

Let  MN  be  the  given  lens,  DOBCF  the  axis  thereof^ 


to  Geometrical  Problems^ 


311 


and  D;z-the  object  whose  image  FH  ^ve  would  find  ;  also 
let  CB  be  the  radius  of  that  surface  of  the  lens  MBN, 
which  v^  nearest  the  object,  and  OB  that  of  the  other 
surface:     make    RC^   perpendicular  to   DF,  and  from 


72,  to  any  point  Eih  the  surface  of  the  lens,  draw  the 

incident  ray  7zE;    and    let    the   continuation  thereof  be 

El,    and  let  the  direction  of   the  same  ray,    after  the 

first  refraction  at  E,  be  E2  ;    and,  after  it  is  refracted  a 

second  time,  at  ^,  let  its  direction  be  e3H  ;    draw  CE 

and  O^R,  and  make  ndv  parallel  to  DF,  calling  O/;,  b ; 

CB,  c;    BD,  d\    Dn^  p ;    and  the  distance  of  the  point 

E  from  the  axis  DF,  :x^ ;    and  let  the  sine  .of  incidence 

be  to  the  sine  of  rviiVaction,    out  of  air  into   glass,  as 

m  to  71.      Then,  the  thickness  of  the  lens  being  looked 

upon  as  inconsiderable   in  respect  of  the  focal  distance 

Fb^  we  shall  have,  as  d  :  x  — p  (E^)  '•  i  d  +  c.  (nv)-  : 

dx  -\-cx  ' —  dp  —  cp         ^         ,  .  ,       ,  ,    ,       ^      .  \     . 
-7—^- ^  =  ^'1  ;  which  added  to  Cv  (/?)  gives 

dx  -^  ex  -—'  cp       ^  ^       ,       ^  dx  -l-  ex  —  cp 

7 =  CI  :  theretore  w  :  n  :  : i- : 

d  d 


72  X  dx  -j-  ex  *—  cf 
dm 


=  C2. 


Moreover,  b  (Ob)  :  x  (BE)  :  : 


b  +  c  (OC)  :  ^  ^  ^X""  ^  CR;   whence  R2  (  =  CR 


C2)r= 


b  -^  c  X  X 


•n  X 


dx  4-  CX' —  cp  ^ 


R2  :  R3  = 


7nx  X  b  -^c      cp  ' 
In         "^ 


dm 
ex  '-^  dx 


but  n  ; 


771 : 


and  therefore 


312  The  Application  of  Algebra 


C3  ('—■RS RCi  —  ^  —  nx  XX  l^  4'C      cp^-^cx^-^dx. 

^"^  ^  '^  hn  d 


Let  X  be  now  taken  =  0,    and  then  C3  will  become 

cp 

-~,  which  let  be  represented  by  C3,  and  draw  B^,  pro- 
ducing the  same  till  it  meets  e3,  produced,  in  H :   then 

^1.        r      r-           r>n\        d -^  C  X  X        m  —  H  X  b  '^- c  X  X 
^3  bemg  (=  C^^  —  C3)  = ^ ^ , 

and    the    triangles   H3^  and   HEB  equiangular,  it  will 

be,  as  (EB  ~^3)  -^ —  :  (B^-)  c  :  : 

nbcd 

(EB)  X  :  BH  =  -=== — ===- — ; =  the  requir- 

m  —  n  X  b  +  c  X  d^-^  nbc 

cd  distance  of   the  image  from    the    lens ;     and    as    c 

(BC)  :  ^  (CG)  :   :  BH  (or  BF)   :  FH  =  ^^^  = 

nbcp 
~  ~  (=  HF)  the  magnitude  or 


m  —  n  X  b  +  c  X  d  -^  7ibc 

the  image,  or  its  linear  amplification. 

COROL.  1. 

Because  the  values  of  BH  and  HF  are  alike  affected 
by  b  and  c,  it  follows  that  both  the  distance  and  magni- 
tude of  the  image  will  remain  unaltered,  if  the  place  of 
the  lens  be  the  same,  let  which  side  you  will  be  turned 
towards  the  object. 

COROL.  2. 

If  d  be  made  infinite,  or  the  distance  of  the  object 

from  the  lens  be  supposed  indefinitely  great,  BF  will  be- 

nbc  ,       .  .        r      1     . 

come  === ,  ;  which  is  the  principal  focal  dis- 

m  —  n  X  b  •}-  c 

tance,  at  which  the  parallel  rays  unite,  and  this  dis- 
tance, when  both  sides  of  the  lens  have  the  same  con- 
vexity, or  ^  is  =  c,  will  become  = :    but  in 

•^'  '  2m  ^-271 

a  planO'ConveXj  where  b  is  infinite,  it  will  be  = 


m  • 


to  Geometrical  Problems, 


and,  in  a  meniscus^  where  b  is  negative,  or  one  surface  con- 
cave, and  the  other  convex,  it  will  be  === . 

m  —  71  X  b  —  c 

The  same  ansxvered  othertvise^  alloxving  also  for  the  thick- 
ness of  the  lens* 

Supposing,  as  before,  that  F  is  the  place  of  the  image 
of  an  object  at  D,  let  FR  and  DS  be  supposed  perpen- 
dicular to  the  axis  FBQ,  intersecting  the  continuation 

R 


of  Ee  (the  intercepted  part  of  the  ray*  DEe'F)  in  r  and 
Sj  and  meeting  the  radii  O^,  CE  (produced)  in  R  and 
S;  likewise  let  Ea  and  ec  be  perpendicular  to  QF,  and 
Ei;  and  Ew  parallel  thereto :  then,  because  the  ray  is 
supposed  to  be  indefinitely  near  the  axis,  ac  may  be 
taken  for  the  thickness  of  the  lens,  which  let  be  de- 
noted by  t ;  putting  ^F  =  2,  ce  -=.1/^  aY,  =  x^  Ob  ,=  ^, 
CB  =  c,  and  BD  =  d  (as  before).  By  similar  tri- 
angles, Ca  (c)  :  aE  (x)  :  :  CD  (c  -f-  ^)  :  DS  =  ^i2SS±J; 
and,  by  the  law  of  refraction,  m  :  n  :  :  DS  :  Ss  *= 
whence    D^    (=    DS  —  Ss)    = 


71  X   X   c  -\-  d 

—    X 

m  c 


n    ^     X   X   c  -\~  d         _  ,  ^  T^. 

1 X — ,    and    vs    ( =  aE   —   D^)  = 

7n  a 

■  2S 


314  The  Application  of  Algebra^  is' c 


X  X  r  —  — ,  by  making  r  =  — ,  and  a'  =  1  —  r. 
c  m 

Now,  vs  :  Ev  (BD)  :  :  «E  t  aQ  (BQ) ;    that  is. 

a:  X  ^  ~  —  :  ^ :  ^  .v  :  ■■  — ^  =  BQ ;  which  is  given 

c  er  —  qd 

from  hence. 

Again,  in  the  very  same  manner,  Oc  (b)  :  ce  (t/)  :  : 

OF  (^  +  z)  •  FH  =  y  X  b  -^  X  ^   ^^^  ^  .  ^  .  .  FR  : 


Rr  =  -^  X  ^       ,  ^— :  whence  Fr  =  1 X  ^       , 

VI  b  mo 

=  ?2!! and  wr  (Fr  —  ce\  -=.  y  x ^5    a^^cl 

b  b 

therefore  cQ  (  =  1  = :  from  which  sub- 

wr  qz  —  br 

tracting  the  value   of   BQ,   found    above,    we  get  this 

equation,  viz. < ,  =  t :  whence  the  va- 

qz  — *  br         cq  —  qd 

cd 

lue  of  2,  by  making  the  given  quantity  t  + =  ^5 

re  —  qd 

vb^ 

comes  out  = :!— -.      But,  if  you  had  rather  have  the 

qg  —  b 

same  in  original  terms,  it  is  but  substituting  for  ^ ;  whence, 

f.           J       ,                              rbcd  4-  rbt  X  re ' —  qd 
alter  reduction,  z  =  ' ^  : 

qd  X  b  +  c  —  rbc  +  qt  X  re  — -  qd 
^vhich,  by  restoring  ?7i  and  72,  becomes 

mnbed  -f  7ipt  x  nc  —  m^-'n  x  d 


m  —  nxmdxb+c — mnbc+m — nxtXnc — m  —  nxd 
where,  if  t  be  taken  =  O,  we  shall  have 

nbcd  ,  r       1 

z  =  -  ',  ■       zr==z ,  the  very  same  as  lound 

m-^72X^X^4-c  —  nbc 
by  the  preceding  method. 


APPENDIX: 

CONTAINING   THE 

CONSTRUCTION 

OF 

GEOMETRICAL  PROBLEMS, 

WITH 

THE   MANNER   OF    RESOLVING    THE    SAME 
NUMERICALLY. 


PROBLEM  L 

The  base^  the  sum  of  the  two  sides^  and  the  angle  at 
:he  vertex  of  any  plane  triangle  being  given;  to  describe 
the  triangle. 

CONSTRUCTION. 

DRAW  the  indefinite  right  line  AE,  in  which 
take  AB  equal  to  the 
sum  of  the  sides,  and 
make  the  angle  ABC  equal  to 
half  the  given  angle  at  the  ver- 
tex, and  upon  the  point  A,  as 
a  center,  with  a  radius  equal 
to  the  given  base,  let  a  circle 
?zCw  be  described,  cutting  BC 
in  C  ;  join  A,  G,  and  make  the 
angle  BCD  =  CBD,  and  let  ^„ 
CD  cut  AB  in  D  ;    then  will  ,  n 

ACD  be  the  triangle  that  was  to  be  constructed. 


316  The  Construction  of 

DEMONSTRATION. 

Because  the  angles  BCD  and  CBD  are  equal,  there- 
fore is  CD  =  DB  (^Euc.  6.  1.),  and  consequently  AD  + 
DC  =  AB  :  likewise,  for  the  same  reason,  the  angle 
AD.C  =  BCD  +  CBD  (Euc.  32.  1.)  is  equal  to  2CBD. 
^.  E.  D. 

Method  of  Calculation. 

In  the  triangle  ABC  are  given  the  two  sides  AB,  AC, 
and  the  angle  ABC,  whence  the  angle  A  is  known  ;  then 
in  the  triangle  ADC  will  be  given  all  the  angles,  and  the 
base  AC  ;  whence  the  sides  AD  and  DC  will  also  be 
known. 

PROBLEM  II. 

The  angle  at  the  vertex^  the  base^  and  the  difference  of 
the  sides ^  being  given  ;  to  determine  the  triangle* 

CONSTRUCTION. 
Draw    AC   at  pleasure,  in  which  take  AD   equal  to 

the  difference  of  the  sides, 
and  make  the  angle  CDB 
equal  to  the  complement 
^  of  half  the  given  angle  to 
a  right  angle  ;  then  from 
the  point  A  draw  AB 
eqitfil  to  the  given  base,  so 
as  to  meet  DB  in  B,  and 
make  the  angle  DBC  = 
CDB,  then  will  ABC  be 
the  triangle  required. 

DEMONSTRATION. 
Since  [by  construction^  the  angles  CDB  and  DBC  are 
equal,  CB  is  equal  to  CD,  and  therefore  CA  —  CB  = 
AD :    moreover,  each  of  those  equal  angles  being  equal 
to  the  complement  of  half  the  given  angle,  their  sum,  which 
is  the  supplement  of  the  angle  C,  must  therefore  be  equal 
to  two  right  angles,  —  the  (whole)  given  angle,  and  con- 
sequently C  =  the  given  angle.     ^.  E,  D, 
Method  of  Calculation, 
In  the  triangle  ABD  are  given  the  sides  AB,  AD, 


Geometrical  Problems. 


5ir 


and  the  angle  ADB  ;  whence  the  angle  A  will  be  given, 
and  consequently  BC  and  AC. 

PROBLEM  III. 

The  angle  at  the  vertex^  the  ratio  of  the  including  sides ^ 
and  either  the  base^  the  perpendicular^  or  difference  of  the 
segments  of  the  base^  being  given  ;  to  describe  the  triangle. 

CONSTRUCTION. 
Draw  CA  at  pleasure,  and  make  the  angle  ACB 
equal  to  the  angle  given ;  take  CB  to  C  A  in  the  given 
ratio  of  the  sides,  and  join  A,  B:  then,  if  the  base  be 
given,  let  AM  be  taken  equal  thereto,  and  draw  ME 
parallel  to  CA  meeting  CB  in  E,  and  make  ED  parallel 
to  AB  ;  but,  if  the  perpendicular  be  given,  let  fall  CF 
perpendicular  tg  AB,  in  which  take  CH  equal  to  the 
given  perpendicular,    and  draw  DHE  parallel  to  AB ; 


^IM  N 


lastly,  if  the  difference  of  the  segments  of  the  base  be  given, 
take  FG  =  AF,  and  join  C,  G,  and  take  GN  equal  to  the 
difference  of  the  segments  given,  drawing  NE  parallel  t^ 
CG,  and  ED  to  BA  (as  before)  ;  then  will  CDE  be^^e 
triangle  which  was  to  be  constructed. 

DEMONSTRATION. 
Because  of  the  parallel  lines  AB,  DE  ;   M^,  AC  ;  and 
NE,  GC  ;  thence  is  DE  =  AM,  and  EI  =^  NG ;  and  also 
CD  :  CE  :  :  CA  :  CB  (iiwc.  4.  6.).     ^,  E.  D. 


318 


The  Construction  of 


Method  of  Calculation. 
Let  AC  be  assumed  at  pleasure  ;  then  the  ratio  of  AC 
to  BC  being  given,  BC  will  become  known ;  and  there- 
fore in  the  xriangle  ACB  will  be  given  two  sides  and  the 
included  angle  ;  whence  the  angles  B  and  A,  or  E  and  D^ 
will  be  found  ;  then  in  the  triangle  EDC,  EHC,  or  EIC. 
according  as  the  base,  perpendicular,  or  the  difference  of 
the  segments  of  the  base  is  given,  you  will  have  one  side 
and  all  the  angles  ;  whence  the  other  sides  will  be  known. 

PROBLEM  IV. 

The  angle  at  the  vertex^  and  the  segments  of  the  base^ 
made  by  a  perpendicular  falling  from  the  said  angle^  being 
given;  to  describe  the  triangle. 

CONSTRUCTION. 

Let  the  given  segments  of  the  base  be  AD  and  DB ; 
■iisect  AB  by  the  perpendicular  EF,  and  make  the  angle 

EBO  equal  to  the  difference 
between  the  given  angle  and 
a  right  one,  and  let  BO  meet 
EF  in  O  ;  from  O,  as  a  cen- 
ter, with  the  radius  OB,  de- 
scribe the  circle  BGAQ,  and 
draw  DC  perpendicular  to 
AB,  meeting  the  periphery  of 
the  circle  in  C  j  join  A,  C,  and 
C,  B,  then  will  ACB  be  the. 
triangle  that  was  to  be  constructed. 

DEMONSTRATION. 
The  angle  ACB,  at  the  periphery,  standing  upon  the 
arch  AQB,  is  equal  to  EOB,  half  the  angle  at  the  cen- 
ter, standing  upon  the  same  arch  ;  but  EBO  is  equal  to 
th»  difference  of  the  given  angle  and  a  right  one  {by  con- 
strutiion)  ;  therefore  ACB  (EOB)  is  equal  to  the  angle 
given.    ^  E.  D. 

Method  of  Cafculation, 
Draw  CFG  parallel  to  AB ;    then  it  will  be,  as  the 
base  AB.  :  the  difference  of  segments  CG  (  :   :   EB  : 
CF)  :  :  the  sine  oi  the  given  angle  at  the  vertex  (EOB)  : 


Peometrkal  Problems^  519 

iRe  sine  of  (COF  =  CBG)  the  diiference  of  the  angles  at 
tlie  base  ;  whence  the  angles  themselves  are  given. 

After  the  same  manner,  a  segment  of  a  circle  may  be 
described  to  contain  a  given  angle,  when  that  angle  is 
greater  than  a  right  one,  if,  instead-of  BO  being  drawn 
above  AB,  it  be  taken  on  the  contrary  side, 

PROBLEM  V. 

Hamng  given  the  base  ^  the  perpendicular  ^and  the  tingle  a^ 
the  vertex  of  any  plane  triangle^  to  construct  the  triangle. 
CONSTRUCTION. 

Upon  AB,  the  given  base  {see  the  preceding  figure)^  \<:^\ 
the  segment  ACGB  of  a  circle  be  described  to  contain  the 
given  angle,  as  in  the  last  problem  ;  take  EF  equal  to  the 
given  perpendicular,  and  draw  FC  parallel  to  AB,  cutting 
the  periphery  of  the  circle  in  C  ;  join  A,  C,  and  B,  C,  and 
the  thing  is  clone ;  the  demonstration  whereof  is  e\^ident 
from  the  last  problem. 

Method  of  Calculation* 

In  the  triangle  EBO  are  given  all  the  angles  and  the 
side  EB  ;  whence  EO  will  be  known,  and  consequently 
OF  (=  DC  —  EO)  ;  then  it  will  be  as  EB  :  OF  :  ': 
the  sine  of  EOB  (the  given  angle  at  the  vertex)  :  the 
sine  of  OCF,  the  complement  of  (COF  or  CBG)  the 
difference  of  the  angles  at  the  base;  whence  these  angles 
themselves  are  likewise  given.  This  calculation  is  adapted 
to  the  logarithmic  canon  ;  but,  by  means  of  a  table  of  na- 
tural sines,  the  same  result  may  be  brought  out  by  one 
proportion  only :  for  BE  being  the  sine  of  BOE,  and  OE 
and  OF  co-sines  of  BOE  and  COF  (answering  to  the 
equal  radii  OB  and  OC),  it  will  therefore  be,  BE  :  EF 
:  :  sine  BOE  ( ACB)  :  co-sine  BOE  +  co-sine  COF  ;  from 
which,  by  subtracting  the  co-sine  of  BOE,  the  co-sin^,  of 
COF  (=  CBG)  is  fouhd. 

PROBLEM  VI. 

The  angle  at  the  vertex^  the  sum  cf  the  txvo  including 
sides ^  and  the  difference  of  the  segments  of  the  base  being 
given^  to  describe  the  triangle. 


02O 


The  Construction  of 


CONSTRUCTION. 

Draw  the  right  line  AC  at  pleasure,  in  which  take 
AB  equal  to  the  difference  of  the  segments  of  the  base, 
and  make  the  angle  CBE  equal  to  half  the  supplement 

of  the  given  angle  ; 
and  from  A  to  BE 
apply  AE  equal  to  ' 
the  given  sum  of  the 
sides ;  make  the  an- 
gle EBD  =  BED, 
and  letBD  meet  AE 
in  D,  and  from  the 
center  D,  with  the 
radius  DB, describe 
the  circle  DEC,  cut- 
ting AC  in  C,  and 
join  D,  C  ;  then  will  ACD  be  the  triangle  required. 
DEMONSTRATION. 
The  angle  EBD  being  -=  BED,  therefore  is  DE  = 
DB  =  DC,  and  consequently  AD  +  DC  =  AE.     More- 
over, the  angle  CDE,  at  the  center,  is  double  to  the  angle 
CBE,  at  the  periphery,  both  standing  upon  the  same  arch 
CE  ;  which  last  (by  construction)  is  equal  to  half  the  sup- 
plement of  the  given  angle  ;  therefore  CDE  is  equal  to  the 
whole  supplement,  and  consequently  ADC  equal  to  the 
given  angle  itself.     ^.  E.  D* 

Method  of  Calculation* 
In  the  triangle  ABE  are  given  the  two  sides  AB,  AE, 
and  the  angle  ABE  ;  whence  the  angle  A  will  be  given ; 
then  in  the  triangle  ABD  will  be  given  all  the  angles  and 
the  side  AB ;  whence  AD  and  DC  (DB)  will  be  also 
given. 

PROBLEM  VII. 

The  angle  at  the  vertex^  the  sum  of  the  including  sides  ^  and 
the  ratio  of  the  segments  of  the  base  being  given  ;  to  deter* 
mine  the  triangle. 

CONSTRUCTION. 

Let  AG  be  to  GB  in  the  given  ratio  of  the  segments 
of  the  base,  and,  upon  the  right  line  AB,  let  a  segment 


Geometrical  Problems.  321 

of  a  6ircle  be  described,  capable  of  containing  the  given 
angle ;  draw  GC  per- 
pendicular to  AB, 
meeting  the  periphe- 
ry in  G ;  join  A,  C, 
afnd  C,  B,  and  in  AC, 
produced,  take  CH 
=  CB  ;  jom  B,  H, 
and  in  HA  take  HD 
equal  to  the  given  sum 
of  the  sidts,  draw  DE 
parallel  to  AB,  and 
EF  to  BC  ;  then  will  DEF  be  the  triangle  required. 

DEMONSTRATION. 

Let  Ftz  be  perpendicular  to  DE.  Whereas  {by  con- 
struction) CH  is  equal  to  CB,  and  FE  parallel  to  CB, 
therefore  is  FE  =  FH  {Euc*  4.  6.)  and  consequently 
FE  ^-  FD  =  HD:  also,  because  FE  is  parallel  to  CB, 
therefore  is  the  angle  DFE  ==  AGB :  moreover,  the 
triangles  ABC,  DEF,  being  equiangular,  it  will  be,  as 
AG  :  GB  :  :  D;i  :  ?zE.     ^E.  B. 

Method  of  Calculation. 

From  the  center  O,  conceive  AO  and  OC  to  be 
drawn ;  supposing  KOI  perpendicular,  and  CI  parallel 
to  AB  :  then  it  will  be,  as  AK  is  to  CI  (KG)  so  is  the 
sine  of  AOK  (=  ACB,  see  prob.  4.)  to  the  sine  of 
COI,  the  difference  of  the  angles  ABC  and  BAC ; 
which  are  both  given  "from  hence,  because  their  sum  is 
given  by  the  question :  therefore  in  the  triangle  DHE 
are  given  all  the  angles  and  the  side  HD,  whence  the  base 
DE  will  be  known. 

PROBLEM  VIIL 

Having  the  angle  at  the  vertex^  the  difference  of  the  in- 
cluding sides ^  arid  the  difference  of  the  segments  of  the  bascj 
to  describe  the  triangle. 

CONSTRUCTION. 

Take  AB  equal  to  the  difference  of  the  segments  of 
the  base,  and  make  tjie  angle  KUn  equal  to  half  the 

2T  ' 


322  The  ConstructioJi  of 

given  angle;    from  A  to  Btz  apply  A E  =  the  difFer- 

ence  of  the  sides ; 
produce  AE,  and 
make  the  angle  EBO 
=  BEO,  and  let  BO 
meet  AE,  produced 
in  O,  and  from  the 
center  O,  at  the  dis^ 
tance  of  OB,  de- 
scribe the  circumfe- 
rence of  a  circle,  cut- 
ting AB  produced  in 

C  J  join  O,  C  ;  then  is  AOC  the  triangle  sought. 

DEMONSTRATION- 

Because  the  angle  EBO  is  =  BEO  {by  construction) ; 
therefore  is  EO  =  BO  =  CO,  and  consequently  AO 
~  OC  =  AE,  Furthermore,  because  the  angle  AOC 
is  double  to  ADC,  and  ADC  =  ABE  {Euc.  corol.  22. 
3.),  therefore  is  AOC  also  double  to  ABE.     %  E.  D. 

Method  of  Calculation. 
The  two  sides  AB,  AE,  and  the  angle  ABE  being 
given,  the  angle  A  will  from  thence  be  found;  then  in  the 
triangle  ABO  will  be  given  all  the  angles  and  the  side 
AB,  whence  OB  (OC)  and  OA  will  be  known. 

PROBLEM  IX. 

The  angle  at  the  vertex^  the  difference  of  the  including 
sides ^  and  the  ratio  of  the  segments  of  the  base^  being  gvoen^ 
■to  determine  the  triangle. 

CONSTRUCTION. 

Let  AG  be  to  GB  in 
the  given  ratio  of  the  seg- 
ments of  the  base,  and  up- 
on the  right  line  AB  let  a 
segment  of  a  circle  ACB 
be  described  {by  prob.  4.} 
_  _^  capable  of   the   given  an- 

A  6  H  gle ;    draw  GC  perpendi- 

cular to  AB,  meeting  the  periphery  in  C,  and  join  A,  C, 


Geometrical  Problems* 


323 


and  B,  C ;  in  AC  take  AP  =  BC,  and  draw  BP ;  also, 
in  AC,  take  CQ  equal  to  the  given  difference  of  the  sides, 
drawing  QE  parallel  to  PB,  and  ED  to  BA  ;  then  will 
CDE  be  the  triangle  which  was  to  be  described. 

DEMONSTRATION. 

The  angle  DCE  is  equal  to  the  given  angle  by  con» 
struction;  also  EQ  being  parallel  to  BP,  DE  to  AB> 
and  AP  =  BC,  therefore  must  DQ  =  EC  {Euc.  4.  6.), 
and  consequently  DC  —  EC  =  CQ.  Moreover,  it  CG 
be  supposed  to  cut  DE  in  tz,  then  D/z :  E/z :  :  AG  :  GB. 
^.  E.  D. 

Method  of  Calculation, 

Let  Cm  be  equal  to  CE,  and  let  Em  be  drawn.  It  will 
be,  as  AB  is  to  AG  —  BG,  so  is  the  sine  of  ACB  to  the 
sine  of  the  difference  of  CBA  and  CAB  {by  prod.  4.) ; 
then  in  the  triangle  DEm  will  be  given  all  the  angles  and 
the  side  Dwz,  whence  DE  will  be  given. 

PROBLEM  X. 

The  angle  at  the  vertex^  the  perpendicular^  and  the  dif- 
ference of  the  segments  of  the  base^  being  given^  to  construct 
the  triangle. 

CONSTRUCTION. 

Draw  RS  at  pleasure,  in  which  take  DE  equal  to  half 
the  difference  of  the  segments  of  the  base,  and  make 
EC  perpendicular  to  RS  and  equal  to  the  given  per- 


pendicular, and  the  angle  DEn  equal  to  the  difference 
between  the  given  angle  and  a  right  one;  join  D,  C, 
and   draw  D;iO  parallel  to  CE,  and  in  DC  take  the 


324 


The  Construction  of 


point  /,  so  that  np  (when  drawn)  may  be  equal  to  7iE  ; 
draw  CO  parallel  to  n/?,  meeting  DnO  in  O ;  and  upon  O 
as  a  center,  with  the  radius  OC,  describe  the  circle  BCA, 
cutting  RS  m  B  and  A ;  join  A,  C,  and  B,  C,  and  the 
thing  is  done. 

DEMONSTRATION. 

Join  O,  B,  and  O,  A  :  since  OC  is  parallel  to  pn^ 
therefore  is  OC  :  DO  \  \  pn  -.  tzD,  or  OB  :  DO  :  : 
nE  :  nD  ;  and  consequently  the  triangle  OBD  similar 
to  the  triangle  ^zED  (hy  Eiic.  7.  6.).  Therefore,  seeing 
the  angle  DE/z  is  {by  construction)  equal  to  the  excess  of 
the  given  angle  above  a  right  one,  ACB  must  be  equal 
to  the  angle  given  [by  prob.  4.).  Moreover,  since  AD 
is  =  DB,  AE  —  BE  w^ill  be  equal  to  2DE,  which  is 
the  given  difference  of  the  segments  (by  construction^, 
^E.D. 

3Iethod  of  Calculation, 

In  the  triangle  CDE,  right-angled  at  E,  are  given  both 
the  legs  DE  and  EC,  whence  the  angle  EDC  will  be 
known,  and  consequently  ODC  ;  then,  as  the  radius  is  to 
the  sine  of  DBO  (:  :  OB  :  DO  :  :  OC  :  DO)  so  is  the  sine 
of  ODC  to  the  sine  of  OCD  ;  whence  DOC,  the  differ- 
ence of  the  angles  ABC,  BAC  (see prob.  4.),  is  also  given, 
and  from  thence  the  angles  themselves. 

PROBLEM  XL 

The  angle  at  the  vertex^  the  perpendicular^  and  the  ratio 
of  the  segments  of  the  base^  being  given^  to  construct  the 
triangle* 

CONSTRUCTION. 

Take  AF  to  FB  in  the 
given  ratio  of  the  seg- 
ments of  the  base,  and 
upon  the  right  line  AB 
describe  a  segment  of  a 
circle  ACB  capable  of 
the  given  angle ;  make 
FC  perpendicular  to  AB, 
ineeting  the  circumference 
of  the  circle  in  C,  in  which 


Geometrical  Problems.  325 

take  CG  equal  to  the  given  perpendicular ;  draw  DGE 
parallel  to  AB,  meeting  AC  and  CB  in  D  and  E ;  and 
then  DCE  will  be  the  triangle  required. 

DEMONSTRATION. 

Because  of  the  parallel  lines  DE  and  AB,  it  will  be  as 
AF  :  DO  (:  :  CF  :  CG)  :  :  FB  :  GE,  or  AF  :  FB  :  : 

DG  :  GE  ;  whence  it  appears,  that  DG  and  GE  are  in  the 
ratio  given.  Also  the  angle  DCE  and  the  perpendicular 
CG  ar.e  respectively  equal  to  the  given  angle  and  perpendi- 
cular, by  construction.     ^.  E.  D. 

Method  of  Calculation. 
As  AB  is  to  AF  —  BF  {see  proh.  4.j,  so  is  the  sine  of 
AC B  to  the  sine  of  the  difference  of  A  and  B;  whence^ 
both  A  and  B  will  be  given,  because  their  sum,  or  the  an- 
gle at  the  vertex,  is  given:  then,  in  the  triangles  DGC, 
EGC,  will  be  given  all  the  angles  and  the  perpendicular 
CG,  whence  the  sides  will  also  be  known. 

PROBLEM  XII. 

The  hase^  the  sum  of  the  sides,  and  the  difference  of  the 
angles  at  the  base^  being  given^  to  describe  the  triangle. 

CONSTRUCTrON. 

At  the  extremity  of  the  base  AB,  erect  the  perpen- 
dicular BE,  and 
make  the  angle 
EBC  equal  to 
half  the  given  dif- 
ference of  the  an- 
gles at  the  base  ; 
from  the  point  A, 
to  BC,  apply  AC 
equal  to  the  sum 
of  the  sides ;  and 
make  the  angle 
CBD  =  BCA; 
then  will  ABD  be  the  triangle  required. 

DEMONSTRATION. 

From  the  centre  D,  with  the  radius  CD,  describe  the 


326  The  Construction  of 

semicircle  CHF,  and  join  F,  B.  Then,  whereas  by  con- 
struction the  angle  CBD  is  =  BCD,  therefore  is  I3B  = 
DC  ;  whence  it  appears  that  AD  -f  Y>1^  is  =  AC,  and 
that  the  semicircle  must  pass  through  the  point  B  :  there- 
fore the  angle  CBF,  standing  in  a  semicircle,  being  a 
right  angle,  and  therefore  =  ABE,  let  FBE,  which  is 
common,  be  taken  away,  and  there  will  remain  ABF 
=  EBC  ;  but  DF  being  equal  to  DB,  it  is  manifest  that 
ABF  (EBC)  is  equal  to  half  the  difference  of  the  angles 
ABD  and  DAB.     %  E.  D. 

Method  of  Calculation. 
As  the  sum  of  the  sides  (AC)  is  to  the  base  (AB),  so 
is  the  sine  of  ABC,  or  of  the  complement  of  half  the 
given  difference,  to  the  sine  of  (C)  half  the  angle  at  the 
vertex  \  whence  the  other  angles  BAD  and  ABD  are  also 
given. 

PROBLEM  XIIL 

The  hase^  the  difference  of  the  sides^  and  the  difference  oj 
the  angles  at  the  base^  being  given^  to  determine  the  triangle* 
CONSTRUCTION. 
At  the  extremity  B  of  the  given  base  AB,  make  the 

angle  ABD  equal 
to  half  the  given 
difference  of  the 
angles  at  the  base ; 
and  from  A  to  BD 
apply  AD  =  the 
difference     of     the 

sides  ;  draw  ADC, 

B  A        and    make  the    an- 

gle  DBC  =  BDC,  and  ABC  will  be  the  triangle  re- 
quired. 

DEMONSTRATION. 
Because  the  angle  DBC  is  =  BDC,  CD  will  be  = 
CB,  and  AC  will  exceed  BC  by  AD.  Moreover,  since 
A  +  ABD  =  (CBD)  CDB  (Euc.  32.  1.),  therefore  is 
A  -f  2ABD  (=  CBD  4-  ABD)  =  ABC,  and  consequent- 
ly ABC  —  A  =  2ABD,  equal  to  the  difference  given. 
4  ^.  D. 


Geometrical  Problems, 


>S27 


Method  of  Calculation. 
In  the  triangle  ABD  are  given  the  two  sides,  AB  and 
AD,    and  the  angle   ABD,^  whence  the  angles  A  and 
ADB  will  be  given,  and  from  thence  the  angles  CBA 
and  ACB. 

PROBLEM  XIV. 

The  difference  of  the  angles  at  the  hase^  the  ratio  of  the 
sides ^  and  either  the  hase^  the  perpendicular^  or  the  difference 
of  the  segments  of  the  base^  being  giveii^  to  describe  the  ti'i- 
angle. 

CONSTRUCTION. 

Draw  AC  at  pleasure,  and  make  the  angle  ACD  equal 
to  the  given  difference  of  the  angles  at  the  base,  and  take 
CD  to  C  A,  in  the  given  ratio  of  the  sides  ;  draw  ADE, 
upon  which  let  fall  the 
perpendicular  CQ;  take 
QE  equal  to  QD,  and 
join  E,  C  ;  then,  if  the 
base  be  given,  let  AB  be 
taken  equal  thereto,  and 
draw  BF  parallel  to  CA 
(meeting  CE  in  F),  and 
FG  parallel  to  E  A  ;  but 
if  the  perpendicular  be 
given,  let  CP  be  taken 
equal  thereto,  and  through  P  draw  FPG  parallel  to  AE  ; 
lastly,  if  the  difference  of  the  segments  of  the  base  be  given^ 
then  let  AR  be  taken  equal  to  that  difference,  draw  RH 
parallel  to  CA,  and  FHG  to  EA  ;  then  will  CFG  be  die 
triangle  required. 

DEMONSTRATION. 
Since  QE  =  QD,  and  the  angle  EQC  =  DQC,  there- 
fore is  CE  =  CD,  and  the  angle  E  =  QDC  =  A  +  ACD 
{Euc.  32.  1.),  and  therefore  E  —  A  =  ACD  ;  whence,  by- 
reason  of  the  parallel  hues  AE,  GF,  £sPc.  we  have  GFC  — 
FGC  ==  ACD,  also  FG  =  AB,  GH  =  AR,  and  CF  : 
CG  :  :  CE  (CD)  :  CA.     .%  E.  D. 


328 


The  Constnictton  of 


Method  of  Calculatzoii, 
Let  CA  and  CD  be  expressed  by  the  numbers  exhibit- 
ing the  given  ratio  of  the  sides  :  then  in  the  triangle  ACD 
will  be  given  two  sides  and  the  included  angle  A  CD  ; 
whence  the  angles  CAE  (CGF)  and  CEA  (CFG)  wiM 
be  given,  and  from  thence  the  sides  CG  and  CF. 


PROBLEM  XV. 

The  hascy  the  perpendiciuar^  and  the  difference  of  the  an- 
gles at  the  hasey  being  given^  to  construct  the  triangle* 

CONSTRUCTION. 

Bisect  the  given  base  AB  by  the  perpendicular  DF, 
In  which  take  DE  equal  to  the  given  height  of  the 
triangle ;     draw  CEGH  parallel  to  AB,  and  make  the 

angle   EDH    equal 
P  to    the    given    dif*- 

ference  of  the  an- 
gles at  the  base ; 
draw  EAQ,  and 
take  Q  therein^ 
so  that  QD  3= 
DH  ;  and,  paral- 
lel to  QD,  draw 
AO,  meeting  DE 
in  O  ;  upon  O,  as 
a  center,  with  the 
radius  OA,  describe  the  circle  AGFCB,  and  from  the 
point  G,  where  it  cuts  the  right  line  CH,  draw  GA  and 
GB  ;  then  Will  AGB  be  the  triangle  required. 
DEMONSTRATION. 
Let  OG  and  BC  be  drawn.  By  reason  of  the  paral- 
lel lines  QD  and  AO,  it  will  be  QD  (DH)  :  AO  (OG) 
3  :  ED  :  EO  ;  therefore  the  two  triangles  EHD,  EGO, 
having  one  angle,  E,  common,  and  the  sides  about  the 
other  angles  D  and  O  proportional,  are  equiangular 
^EuG.  7.  6.),  and  consequently  EOCt  =  EDH.  More- 
over, because  DOEF  is  perpendicular  both  to  AB  and 
GC,  and  AD  equal  to  BD,  it  is  evident  that  the  circle 
passes  through  the  point  B,  and  that  the  arches  FC,  FCt, 


Geometrical  Problems. 


329 


as  well  as  the  angles  ABC,  BAG,  are  equal ;  and  con- 
sequently that  the  angle  GBC  is  the  difference  of  the  an- 
gles BAG,  ABG :  but  this  difference  GBC  is  equal  to 
EGG,  or  EDH  (Euc.  20.  3.),  that  is,  equal  to  the  differ- 
ence given.     ^.  JE.  D. 

Method  of  Calculation. 
First,  in  the  right-angled  triangle  AED  are  given  both 
the  legs  AD  and  DE,  whence  the  angle  DEA  will  be 
given :  then  it  will  be,  as  the  radius  is  to  the  sine  of  the 
angle  H,  the  complement  of  the  given  difference  (  :  :  DH 
:  DE  :  :  DQ  :  DE),  so  is  the  sine  of  DEA  to  the  sinejof 
Q ;  whence  AGE  (QDE)  will  also  be  given ;  from 
which  take  GGE,  and  there  will  remain  AGG,  equal  to 
twice  ABG,  the  lesser  angle  at  the  base. 

PROBLEM  XVI. 


The  sum  of  the  sides^  the  difference  of  the  segments  of 
the  base^  and  the  diff'erence  of  the  angles  at  the  base^  being 
given^  to  describe  the  triangle. 

CONSTRUCTION. 

Make  AD  equal  to  the  sum  of  the  sides,  and  the 
angle  ADE  equal  to  half  the  difference  of  the  angles 
at  the  base  ;  from  A  to 
DE  apply  AE  equal  to 
the  given  difference  of 
the  segments  of  the  base; 
make  the  angle  CED 
=  EDC,  and  from  the 
point  C,  where  EC  cuts 
AD,  with  the  radius  EC, 
describe  the  semicircle 
FEB,  cutting  AE,  produced  in  B  ;  join  B,  C,  and  the 
thing  is  done. 

DEMONSTRATION. 

Upon  AB  let  fall  the  perpendicular  CQ. 

Because  EQ  is  =  BQ  {Etic.  3.  3.),  therefore  will  AQ 
—  BQ  =  AE  :  also,  because  the  angles  CED,  EDC 
are  equal  {by  construction)^  CD  will  be  =  CE  =  CB, 
and  consequently  AC  -f  CB  =  AD.     Moreover,  AJftC 

2U 


33a 


The  Construction  of 

(i:«c.  32.  l.)=:2ADE 


w  BAG  =:  BEC  —  BAG  =  AGE 

(Euc.  20.  3.)     ^  E.  D. 

Method  of  Calculation. 
In  the  triangle  ADE  are  given  the  sides  AD,  AE,  and 
the  angle  D,  whence  the  angle  A  will  be  given ;  then,  in 
the  triangle  AGE  are  given  all  the  angles,  and  the  side 
AE,  whence  AG  and  GB  (GE)  will  be  likewise  given.! 

PROBLEM  XVII. 

The  difference  of  the  angles  at  the  base^  the  ratio  of  the 
segments  of  the  base^  and  either  the  sum  of  the  sides ^  the 
difference  of  the  sides^  or  the  perpendicular^  being  given^ 
to  construct  the  triangle. 

GONSTRUGTION. 

Let  AG  be  to  BG  in  the  given  ratio  of  the  segments 
of  the  base ;  and  upon  AB  let  a  segment  of  a  circle 
BPA  be  described  (by  problem  4),  to  contain  an  angle 

equal  to  the  difference 
of  the  angles  at  the 
base ;  raise  GP  per- 
pendicular to  AG,  cut- 
ting the  periphery  of 
the  circle  in  P,  and  in 
AG  produced,  take  GD 
=  GB,  and  draw  PA, 
PB  and  PD  :  then,  if 
the  perpendicular  be 
given,  take  PF  equal 
thereto,  and  through  F 
draw  EFG  parallel  to  AD  ;  but  if  the  sum  or  difference  of 
the  sides  be  given,  let  a  fourth  proportional  PE,  to  AP  ± 
PD,  AP,  and  the  said  sum  or  difference  be  taken,  and  draw 
EFG  as  above  ;  then  will  PEG  be  the  triangle  required. 
DEMONSTRATION. 
Since  GP  is  perpendicular  to  AD,  and  GD  =  GB,  the 
angle  D  will  be  equal  to  DBP  =  A  +  BPA :  whence, 
because  EG  is  parallel  to  AD,  PGE  will  be  =  PEG 
•Ar  BPA  {Euc.  29.  lO^  and  consequently  PGE  —  PEG 


Geometrical  Problems. 


331 


1=1  ABP,  whicU,  by  construction,  is  equal  to  the  given 
difference  of  the  angles  at  the  base. 

J  Again,  by  reason  of  the  parallel  lines  AD  and  EG, 
it  will  be,  EF  :  FG  :  :  AC  :  (BC)  CD.  Likewise,  for 
the  same  reason,  AP  ±  PD  :  PA  :  :  PE  ±  PG  :  PE  :  : 
given  sum  or  diff.  of  sides  :  PE  (by  constniction)  and 
consequently  PE  ±  PG  =  the  said  given  sum  or  differ- 
ence,    i^.  E.  Z). 

Meth  od  of  Calculati  o  n . 

First,  it  will  be,  as  AB  is  to  AD,  so  is  the  sine  of  APB 
to  .the  sine  of  APD  {by  proh,  4)  ;  then,  in  the  triangle 
PGE,  will  be  given  all  the  angles,  and  either  the  perpen- 
dicular, or  the  sum  or  difference  of  the  sides,  whence  the 
sides  themselves  are  readily  determined. 

Note.  The  perpendicular  cutting  the  circle  in  two  points, 
indicates  that  this  problem  is  capable  of  two  differeixt  so- 
lutions* 

PROBLEM  XVIIL 

The  difference  of  the  sides:  ^  the  dijference  of  the  sej^mcnts 
'ofthe  base^  and  the  difference  of  the  angles  at  the  basc^  be-' 
ing  given^  to  describe  the  triangle. 

CONSTRUCTION. 

Draw  the  indefinite  line  AQ,  in  which  take  AD 
equal  to  the  given  difference  of  the  sides,  and  make  the 
angle  QDH  e- 
qual  to  the  com- 
plement of  half 
the  difference  of 
the  angles  at  the 
base ;  from  A  to 
DH  apply  AC 
=  the  given  dif- 
ference of  the 
segments  ;  and 
having  produced 
the  same  to  L, 
make  the  angle  DCE  equal  to  CDE,  and  let  CE  meet  AQ 
in  E,  and  upon  the  center  E,  with  the  radius  EC,  describe 
an  arch,  cutting  AL  in  B  ;  join  E',  B,  so  shall  AEB  be  the 
triangle  required. 


332 


The  Construction  of 


DEMONSTRATION. 

Upon  AB  let  fall  the  perpendicular  EP. 

Because  the  angle  DCE  =  CDE,  therefore  is  ED 
=  EC,  and  consequently  AE  —  EB  (=  AE  - —  EC  = 
AE  —  ED)  ==  AD.  Also,  since  EB  =  EC,  therefore 
will  PB  z=z  PC,  and  consequently  AP  —  BP  (AP  —  PC) 
r=  AC.  Moreover,  the  angle  EBC  being  =  ECB  {Euc. 
5. 1.),  and  ECB  —  A  ==  CE  A  {Euc.  32.  1.),  it  is  plain  that 
EBC  —  A  =  CEA,  equal  to  the  given  difference,  because 
the  triangle  EDC  is  isosceles,  and  the  angle  at  the  base 
equal  to  the  complement  of  half  the  said  difference,  by 
construction.     ^.  E.  JD. 

Method  of  Calculation. 

In  the  triangle  ADC  are  given  two  sides,  and  the  angle 
ADC,  whence  the  angle  A  will  be  kno^vn ;  then,  in  the 
triangle  ACE,  will  be  given  all  the  angles  and  the  side  AC, 
whence  AE  and  CE  (BE)  will  also  become  known, 

PROBLEM  XIX. 


The  perpendicular^  the  difference  of  the  angles  at  the  base^ 
(ind  the  difference  of  the  segments  of  the  hase^  being  given ^ 
to  construct  the  triangle, 

CONSTRUCTION. 

Upon  AQ,  equal  to  the  given  difference  of  the  seg- 
ments of  the'  base,  let  a  segment  of  a  circle  QCA  be 
described,  capable  of  the  difference  of  the  angles  at  the 

base  ;  bisect  AQ  with 
the  perpendicular  TL, 
in  which  let  TE  be 
taken  equal  to  the 
given  perpendicular  ; 
draw  EC  parallel  to 
AQ,  cutting  the  pe- 
riphery of  the  circle 
in  C  \  also  draw  CP 
perpendicular  to  AQ, 
and  in  AQ  produced 
take  PB  =  PQ ;  join 
Cj  A|^nd  C,  B  j  then  will  ACB  be  the  triangle  required. 


Geometrical  Problems. 


335 


DEMONSTRATION. 

Since  {by  construction)  CP  is  perpendicular  to  QB, 
and  PB  equal  to  PQ,  thence  will  the  angle  B  =  PQC, 
and  B  (PQC)  —  BAG  =  ACQ  =  difference  of  an- 
gles given  :  also,  for  the  same  reason,  will  CP  =  TE,  and 
AP  —  BP  =  AP  —  PQ  =  AQ.  4  ^-  ^• 
Method  of  Calculation. 

From  the  center  O,  conceive  O  A  and  OC  to  be  drawn : 
then,  in  the  triangle  AOT,  will  be  given  all  the  angles, 
and  the  side  AT,  whence  OT  and  OE  will  be  given  ; 
then  it  will  be,  as  AT  :  OE  :  :  sine  of  AOT  (ACQ)  : 
sine  of  OCE  ;  whence  all  the  angles  in  the  figure  are 
given. 

PROBLEM  XX. 


The  segments  of  the  base^  and  the  sum  of  the  sides  of  any 
plane  triangle^  being  ^iven^  to  determine  the  triangle. 
CONSTRUCTION. 

From  the  greater  segment  AQ,  take  QF  equal  to  the 
lesser  segment   BQ ;     make   QL  perpendicular  to  AB, 
and  draw  AI,  mak- 
ing any  angle  with 
AB  at  pleasure,  in 
which  take  AE  e- 
qual  to    the    given 
sum    of  the  sides, 
and   join     B,     E  ; 
make     the      angle 
AFG  =  AEB,and 
bisect    EG    in    H, 
and    from  B    as  a 
center,  with  the  ra- 
dius EH,  describe  tjiC/z,  cutting  the  perpendicular  QL  in 
C  \  join  C,  A,  and  C,  B,  and  the  thing  is  done. 
DEMONSTRATION. 

From  the  center  C,  with  the  radius  CB,  let  the  circle 
BDLKF  be  described  ;  and  let  AC  be  produced  to  meet 
its  periphery  in  D.  By  reason  of  the  similar  triangles 
AEB,  AFG,  it  will  be,  as  AE  :  AB  :  :  AF  :  AG; 
whence  AG  X  AE  =  AF  X  AB  ;    but  {by  Euc.  37.  3.) 


,334  The  Construction  of 

AF  X  AB  =  AK  X  AD  ;  therefore  is  AG  X  AE  =  AK 
X  AD  :  whence,  as  EG  and  DK  are  equal,  by  construc- 
tion, it  is  evident  that  AG  and  AK,  as  well  as  AE  and 
AD,  must  be  equal.     ^  E.  D. 

Method  of  Calculation, 
As  AE  :  AB  :  :  AF  :  AG ;  which  taken  from  AE, 
and  the  remainder  divided  by  2,  gives  BC  (EH),  the  lesser 
side  of  the  triangle. 

PROBLEM  XXI. 

The  segments  of  the  base^  and  the  difference  of  the  sides^ 
being'  given  J  to  describe  the  tj'iangle. 

CONSTRUCTION. 

Take  AF  equal  to  the  difference  of  the  given  segments 
AQ,  BQ  (^see  the  preceding  figure)^  and  draw  AI  making 
any  angle  with  AB  at  pleasure,  in  which  take  AG  equal 
to  the  given  difference  of  the  sides  ;  join  F,  G,  and  make 
the  angle  ABE  =  AGF,  and  from  the  center  B,  at  the 
distance  of  -jEG,  describe  nCm^  cutting  the  perpendicular 
QL  in  C  ;  join  C,  B,  and  C,  A,  then  will  ACB  be  the  tri- 
angle that  was  to  be  constructed  :  the  demonstration  of 
which  is  so  very  little  different  from  the  foregoing,  that  it 
would  be  needless  to  give  it  here. 

LEMMA. 

If  a  given  right  line  AB  be  divided  in  any  given  ratio^ 
€tt  C,  and  the  right  line  (ZT^O  he  taken  to  A.C  i?i  the  ratio 
qfBC  to  AC  — ^  BC  ;  andfroin  O  as  a  center^  at  the  distance 
oyTOC,  a  circle  CPD  be  described^  and  txvo  right  lilies  AP, 
BP  be  drawn  from  A  and  B,  to  meet  any  -where  in  the  pe- 
riphery thereof;  I  say  these  lines  will  be  to  one  another 
{every  where)  in  the  given  ratio  of  K^  to  H'Si, 

For, [since  CO  :  AC  :  :  BC  :  AC  — BC;  therefore, 
by  composition,  CO  :  AO  :  :  BC  :  AC  ;  and,  by  per- 
mutation, CO  :  BC  :  :  AO  :  AC;  whence,  by  di- 
vision, CO  :  BO  :  :   AO  :  CO,  or  PO  :  BO  :  :  AO  : 

PO  ;    wherefore,  seeing  the  sides  of  the  triangles  FOB, 
AOP,    about  the  common    angle   O,    are    proportional. 


Geometrical  Frobtems. 


335 


tllose  triangles  must  be  similar  {^Euc.  6.  6.),  and  there- 
fore the  other  sides  also  proportional,  thiat  is,  PO  (CO)  i 


AO  :  :  BP  :  AP :   whence  {by  the  second  step)  BC  :  AC 
::BP;AP.     §>.  E.  D. 


PROBLEM  XXII. 

The  segments  of  the  base^  and  the  ratio  of  the  sides^  being 
given^  to  determine  the  triangle. 

CONSTRUCTION. 

Let  AQ  and  QB  be  the  segments  of  the  base ;  and 
let  the  whole  base  AB  be  divided  at  C,  in  the  given 
ratio  of  the  sides ; 
take  CO  to  AC, 
as  BC  to  AC  — 
BC,  and  with  the 
radius  CO  de- 
scribe the  circle 
CPD,  and  raise 
QP  perpendicu- 
lar to  AO,  meeting  the  periphery  in  P  ;  join  A,  P,  and  B, 
P  ;  then  will  ABP  be  the  triangle  required  :  the  demon- 
stration of  which  is  manifest  from  the  preceding  lemma. 
Method  ofCalculatioiu 

Since  the  ratio  of  AC  to  CB,  and  the  length  of  the 
whole    line    AB    are    given,    thence  will  AC  and  CB 

be  given,    and    consequently  OC    (-— -— ]  ;    from 


336 


The  Construction  of 


O  Q 


whence    the    perpendicular    PQ    ( =  VCQ  X   DQ)    is 

likewise  given. 

PROBLEM  XXIII. 

Having  the  hascy  the  perpendicular^  and  the  ratio  of  tht 
sides^  to  describe  the  triangle* 

CONSTRUCTION. 

Let  the  base  AB  be  divided  at  C,  in  the  given  ratio 
of  the  sides,  and  let  the  circle  CPD  be  described  as  in 

the  last  problem  ; 
^  in     OR,     perpen- 

-^=^=J5^  dicular     to     AD, 

take  On  equal  to 
the  given  perpendi- 
cular, and,  through 
72,  draw  PnP  pa- 
\  ^  rallel  to  AD,  cut- 
ting the  periphery 
of  the  circle  in  P  ;  join  P,  A,  and  P,  B,  and, the  thing  is 
done.  The  truth  of  this  is  also  evident  from  the  preceding 
lemma. 

Method  of  Calculation. 
Upon   AD  let  fall  the  perpendicular    PQ,    and  join 

O,  P  :  then  PO   (  =  -— -^ --—  )   will  be  dven  ;    there- 

\       AC  —  BC/  ^ 

fore,  in  the  triangle  OPQ,  are  given  OP  and  PQ,  from 

whence  not  only  OQ,  but  AQ  and  BQ  are  also  given. 

Note*  The  parallel  P;zP,  cutting  the  circle  in  two  points, 

shows  that  this  problem  admits  of  two  different  solutions. 

PROBLEM  XXIV. 

The  difference  of  the  segments  of  the  base^  the  perpendi- 
cular^ and  the  ratio  of  the  sides^  being  given^  to  construct 
the  triangle. 

CONSTRUCTION. 

Let  AB  be  the  difference  of  the  segments  of  the  base 
{see  the  last  figure^  ^  and  let  every  thing  be  done  as  in 
the  preceding  problem :  take  Q(b  =  QB,  and  join  P,  b  ; 
then  will  A^P  be  the  triangle  required.     The  reasons  of 


Geometrical  Proplems. 


337 


which  are  obvious  from  what  has  been  said  already; 
and  the  numerical  solution  is  also  evident  from  the  last 
problem, 

PROBLEM  XXV. 

The  ratio  of  the  segments  of  the  hase^  the  perpendicular^ 
and  the  ratio  of  the  sides  being  given ^  to  construct  the  tri^ 
anq-le. 

CONSTRUCTION. 

Draw  any  right  line  ABC  at  pleasure,  in  which  take 
AE  to  EB  in  the  given  ratio  of  the  sides,  and  AF  to 


Elf    B    5: 


FB  in  the  given  ratio  of  the  segments  of  the  base,  and 
make  FQ  perpendicular  to  AB,  and  equal  to  the  given 
height  of  the  triangle ;  make  also  EC  :  AE  :  :  BE  :  AE  — 
BE,  and  with  the  radius  CE  describe  the  circle  ERS,  and 
from  the  point  R,  where  it  intersects  the  perpendicular 
FQ,  draw  RA  and  RB,  and  draw  QP  and  QT  parallel  to 
RA  and  RB  ;  then  will  PQT  be  the  triangle  that  was  to 
be  described, 

DEMONSTRATION. 

By  the  foregoing  lemma,  AR  :  BR  :  :  AE  :  BE  ;  there- 
fore, by  reason  of  the  parallel  lines,  it  will  be  QP  :  QT 
{:  :  RA  :  RB)  :  :  AE  :  BE.  And,  for  the  same  reason, 
PF  :  TF  :  :  AF  :  BF.     ^  E.  D. 

Method  of  Calculation, 

Having  assumed  AB  at  pleasure,  there  will  be  given 

BE,  AE,  BF,  and  CE   f=  A|J1_||)  ;    whence  RF 

\      AE  —  BE/ 

(  =  \EF  X  CE  +  CF)  is  also  given  ;  then,  in  the 
right-angled  triangle  BRF,  will  be  given  both  the  legs 

2X 


338 


The  Construction  of 


BF  and  RF,  whence  the  angle  B  is  given ;  lastly,  in  the 
triangle  FQT  will  be  given  all  the  angles  and  the  side 
FQ,  whence  QT  and  TF  will  be  given,  and  consequently 
PQ  and  FP- 

PROBLEM  XXVI. 

To  divide  a  given  angle  ABC  into  twoparts^  CBF,  ABF, 

so  that  their  sines  may  obtain  a  given  ratio. 
CONSTRUCTION. 
In  BA,  and  CB  produced,  take  BE  and  BD  in   the 

given  ratio  of  the  sine  of  CBF  to  the  sine  of  ABF  ; 

draw  DE,  and  parallel 
thereto  draw  BF,  and 
the  thing  is  done.  For, 
by  trigonometry^  BE  : 
BD  :  :  the  sine  of  D 
(=  CBF)  :  the  sine  of 
BED  (=  ABF).  Hence 
the  numerical  solution  is 
also  evident ;  since  it 
will  be,  as  the  sum  of 

BE  and  BD  is  to  their  difference,  so  is  the  tangent  of  half 

the  given  angle  ACB  to  the  tangent  of  half  the  difference 

of  the  two  required  parts  FBC  and  FBA. 

PROBLEM  XXVIL 

To  divide  an  angle  given  into  two  parts ^  so  that  their  tan- 
gents may  be  to  each  other  in  a  given  ratio. 
CONSTRUCTION. 

Take  any  two  right  lines  AD,  BD,  which  are  in  the 
ratio  given,  and  upon  the  whole 
compounded  line  A  B  let  a  seg- 
ment of  a  circle  BCA  be  de- 
scribed, capable  of  the  angle 
given  J  make  DC  perpendicu- 
lar to  AB,  meeting  the  peri- 
phery in  C,  and  draw  AC  and 
BC,  then  will  ACD  and  BCD 
be    the    two     angles    required. 


Geometrical  Problems.  339 

The  reason  of  which  is  evident,  at  one  view,  from  the  con- 
struction. The  method  of  solution  is  also  very  easy  ;  for 
it  will  be,  as  AB  is  to  AD  —  DB,  so  is  the  sine  of  ACB 
to  the  sine  of  B  —  A  {see  problem  4)  ;  whence  B  and  A, 
and  also  BCD  and  ACD,  are  given. 

PROBLEM  XXVIII. 

To  divide  a  given  angle  ABC  into  two  parts^  so  that 
their  secants  may  obtain  a  given  ratio. 

CONSTRUCTION. 

Take  BE  to  BT  in  the  given  ratio  of  the  secants  ; 
jt)in  T,  E,  and  let  BF  be  drawn  perpendicular  to  ET, 


and  the  thing  is  done.     The  truth  of  which  is  manifest 
from  the  construction. 

Method  of  Calculation. 
The  angle  EBT  and  the  ratio  of  the  sides  BE  and  BT 
being  given,  the  angles  E  and  T  will  also  be  given,  and 
consequently  their  complements  EBF  and  FBT. 

PROBLEM  XXIX. 

From  a  given  point  O^  to  drat^  a  right  line  OF,  to  cnt 
two  right  lines  AC,  AB,  given  by  position^  so  that  the 
parts  thereof  OE,  OF,  intercepted  between  that  point  and 
those  lines^  may  be  to  one  another  in  a  given  ratio. 

CONSTRUCTION. 

From  O,  through  A,  the  point  of  concourse  of  BA 
and  CA,  let  OAD  be  drawn,  in  which  take  AD  to  AO 


340  The  Construction  of 

in  the  given  ratio  of  FE  to  EO,  and  draw  DF  parallel  to 


Ac,  cutting  AB  in  F  ;  join  F,  O,  and  the  thing  is  done  ; 
as  is  manifest  from  Euc.  2.  6. 

Method  of  Calculation. 
Since  the  point  O  and  the  lines  AC  and  AB  are  given 
by  position,  OA  and  all  the  angles  at  the  point  A  are 
given;  therefore,  from  the  given  ratio  of  AD  and  AG, 
AD  will  be  likewise  given  ;  then  in  the  triangle  DAF 
will  be  given  AD  and  .all  the  angles  (because  FDA  = 
C  AO)  ',  whence  AF  is  also  given. 

PROBLEM  XXX. 

To  divide  a  given  arch  CD  i7ito  two  such  parts ^  that 
the  rectangle  under  their  sines  may  be  of  a  given  magni- 
tude* 

CONSTRUCTION. 
Upon  the  radius  OC  let  fall  the  perpendicular  DF, 

in    which     (pro- 
JD  duced  if  need  be) 

take  FG  =  |OC, 
and  thereon  con- 
stitute a  rectangle 
FIHG  equal  to 
the  given  rect- 
angle ;  and,  sup- 
posing HI  to  cut 
the  circumfer- 
ence in  E,  draw  OB  to  bisect  DE  j  then  will  CB  and  DB 
be  the  parts  required. 


Geometrical  Problems.  341 

DEMONSTRATION. 

Draw  CM  and  DNE  perpendicular  to  the  radius  OB, 
and  Nti  and  E^  perpendicular  to  DF. 

It  is  evident,  by  construction,  that  the  triangles  OCM 
and  DN^i  are  similar  (because  Nn  is  parallel  to  CO,  and 
IS^D  to  CM)  ;  therefore  OC  :  CM  :  :  DN  :  N/z  (=  \Ee\ 
and  consequently  CM  X  DN  =  OC  X  |E^  =  ^OC  X 
E^  =  FGxEe  =  the  given  rectangle  by  construction. 
^.  E.  D. 

Method  of  Calculation. 

Dividing  the  measure  of  the  given  rectangle  by  half 
the  radius,  FI  will  be  given,  which,  added  to  OF,  the 
co-sine  of  CD,  gives  the  co-sine  (OI)  of  CE,  the  differ- 
ence of  the  two  parts  ;  whence  the  parts  themselves  will 
be  known. 

PROBLEM  XXXI. 

Having  the  ratio  of  the  sines ^  and  the  ratio  of  the  tan- 
gents of  two  angles^  to  determine  the  angles. 

CONSTRUCTION. 

Let  AD  be  to  ED  in  the  given  ratio  of  the  sines,  and 
AD  to  FD  in  the  given  ratio  of  the  tangents  j  and  about 
the  center  D,  with  the  interval  DE,  let  the  semicircle 


ERK  be  described ;  and,  upon  AF,  describe  another 
semicircle,  cutting  the  former  in  H,  and  through  H  draw 
AR,  and  join  H,  D ;  then  will  DHR  and  DAR  be  the 
two  angles  required. 

DEMONSTRATION. 

Join  F,  H,  and  draw  DQ  perpendicular  to  AR. 

The  angle   AHF,  standing  in  a  semicircle,    being   a 
right  one,  the  lines  FH  and  DQare  parallel  (l)yEiic.27.  1.)  ; 


342  The  Construction  of 

and  therefore  AD  :  FD  :  :  AQ  :  HQ  :  :  tang,  of  DHQ  : 
tang.  D AQ.  Likewise  DA  :  DE  (DH)  :  :  sine  of  DHQ  : 
sine  of  D  AQ,  as  was  to  be  shown. 

Method  of  Calculation. 
If  AR  be  supposed  to  meet  the  periphery  in  R,  and 
RN  be  drawn  parallel  to  HF,  meeting  AK  in  N  ;  then 
will  DN  =  DF,  and  AN  :  AR  :  :  AF  :  AH  ;  but 
{by  Euc.  37.  3.)  AR  :  AK  :  :  AE  :  AH  ;  whence,  by 
compounding  the  terms  of  these  two  proportions,  ^c. 
AN  :  AK  :  :  AF  X  AE  :  AH^  ;  whence  AH,  as  well 
as  AD  and  DH,  being  known,  the  angles  A  and  K  will 
also  be  known. 

PROBLEM  XXXIL 

To  draw  from  a  point  A,  in  the  circumference  of  a  given 
circle^  two  subtenses  AB  and  AD,  which  shall  be  to  one 
another  in  the  given  ratio  ofm  to  72,  and  cut  ojftwo  arches 
AB  and  ABD,  in  the  ratio  of\  to  S, 

CONSTRUCTION. 

Draw  the  diameter  AH, 
and  take  the  subtense  AQ, 
in    proportion    thereto,     as 
n  —  m  to  2m ;     from  the 
center    O  draw  OB  paral- 
lel   to    AQ,     meeting    the 
periphery  in    B ;    join    A, 
B,  and  make  the  subtenses 
BC    and    CD    each    equal 
to  AB,  and  draw  AD,  and 
the  thing  is  done. 
DEMONSTRATION. 
Join  H,  Q,  and  draw  BE  and  CF  perpendicular  to 
AD. 

The  angle  AOB  (QAH)  at  the  center,  standing  up- 
on the  arch  AB,  is  equal  to  the  angle  BAD  at  the  cir- 
cumference, standing  upon  double  that  arch ;  therefore, 
AQH  being  equal  to  AEB,  or  a  right  angle  {Euc*  31.  3.), 
the  triangles  AQH,  AEB,  must  be  equiangular,  and 
consequently  AB  :  AE   :  :  AH  :   AQ  ;     but,  by  con- 


Geometrical  Problems* 


343 


struction,  AH  :  AQ  i  \2m  :  n  —  my  whence  AB  :  AE 
:  :  2m:  n  —  w,  or  AB  :  2AE  :  :  2;?z  :  2n  -—  2?w  ;  there- 
fore (^by  composition)  AB  :  AB  +  2AE  (:  :  2m  :  2n)  :  : 
m  :  n.  But  AB  being  =  BC  =  CD,  EF  is  =  BC  =  AB, 
DF  =  AE,  and  AD  =  2AE  +  AB,  Hence  AB  :  AD 
::m:  n.     %  E.  D. 

Method  of  Calculation. 
Let  AP  be  perpendicular  to  OB  ;    then,  because  of 
the  similar  triangles   OAP,  AHQ,  it  will  be  as  AO  : 
OP  (:  :  AH  :  AQ)  :  :  2m  i  n*—  m  (by  construction)  ; 


therefore  OP 


n  —  m  X  AO 

'2m  ' 


(BP  =  AO  ~  OP)  = 


Sm- 


n  X  AO 


2m 


,  and  consequently  AB  (V2AO  X  BP)  = 


4 


Sm  —  n 


X  AO  ;  whence  AD  is  also  given. 


PROBLEM  XXXIIL 

The  area  and  hypothenuse  of  any  right-angled  plane  tri- 
angle being  given^  to  describe  the  triangle. 

a 

CONSTRUCTION. 

Upon  the  given  hypothenuse  AB,  as  a  diameter,  let 
the  semicircle  ACB  be  described,  and  upon  OB,  equal 
to  half  AB  (^by  Euc.  41.  1.),  constitute  the  rectangle  OE 


equal  to  the  given  area  of  the  triangle,  and  let  the  side 
thereof,  EF,  cut  the  periphery  of  the  circle  in  C  ;  join  A, 
C,  and  B,  C,  and  the  thing  is  done. 


o44  The  Constructio7i  of 

DEMONSTRATION. 

The  triangle  ABC,  standing  upon  the  whole  diameter 
AB,  is  equal  to  the  rectangle  OE,  of  the  same  altitude, 
standing  upon  half  AB  {by  Euc.  41.  1.),  which  last  (by 
constructwri)  is  equal  to  the  area  given. 
31ethod  of  Calculation* 

Join  O,  C,  and  let  CD  be  perpendicular  to  AB  ;  then 
it  will  be,  as  A02  (AO  x  OC)  :  AO  x  DC  ( :  :  OC  : 
DC)  :  :  radius  :  sine  of  DOC  ;  which,  in  words,  gives 
this  theorem. 

As  the  square  of  half  the  hypotlieniise  of  any  right-an- 
gled plane  triangle  is  to  the  area^  so  is  the  radius  to  the 
sine  of  double  the  lesser  of  the  two  acute  angles* 

N*  B.  Since  no  sine  can  be  greater  than  the  radius,  it 
is  plain,  that,  if  the  square  of  half  the  hypothenuse  be  not 
given  greater  than  the  area  of  the  triangle,  the  problem 
will  become  impossible  ;  in  which  case  the  side  EF,  in- 
stead of  cutting,  will  pass  quite  above  the  circle. 

PROBLEM  XXXIV. 

To  describe  a  right-angled  triangle^  whose  area  shall  be 
equal  to  a  given  square^  and  the  sum  of  its  two  legs  equal  to 
a  given  right  line  AB. 

CONSTRUCTION. 

Upon  AB  let  a  semicircle  be  described ;  make  ACD 

=  half  a  right  an- 
gle, and  CD  =  twice 
(PQ)  the  side  of  the 
given  square  ;  draw 
Iq  DE  parallel  to  AB, 
meeting  the  circum- 
ference in  E,  and  EF 
perpendicular  to  AB, 
p  intersecting  AB   in  F, 

in  which,  produced, 
take  FG  ~  FB,  and  draw  AG  ;  so  shall  AFG  be  the  tri- 
anjrle  required* 

DEMONSTRATION. 
It  is  evident  that  AF  +  FG  =  AB  ;  and  also  that  the 


P 


Geometrical  Problems*  3^45 

area  AFG  =  |AF  x  FG  =  |AF  x  FB  =  |FE^  (  = 
^DH^)  =  1CD2  =  PQ2.     ^  E.  D. 

Method  of  Calculation. 
If  the  radius  CE   be  drawn,  in  the  right-angled  trl- 
angle  CEF,  there  will  be  given  CE  (  r=  ^AB)  and  EF]* 
( =  2PQ2 j  whence  CF  (  =  V^AB^  —  2PQ2)  will  be 
known,  and,  from  thence,  both  AF  and  FG. 

LEMMA. 

The  area  of  any  right-angled  triangle^  ABC,  is  equal 
to  a  rectangle  under  half  its  perimeter  and  the  excess  of  that 
half  perimeter  above  the  hypothenuse^  or  longest  side. . 

DEMONSTRATION. 

In  the  proposed  triangle  let  the  circle  EGF  be  in- 
scribed, and  from  the  center  D,  to  the  angular  points 
A,  B,  C,  and  the  points  of  contact  E,  F,  G,  let  the 
right  lines  DA,  DB,  DC,  DE,  DF,  and  DG  be  drawn. 
It  is  plain  that  the  sum  of  the  three  triangles  ADB, 
BDC,  and  ADC,  is 

equal  to  the   whole  y/\(^ 

triangle  ABC  ;  but 
the  triangle  ADB  is 
equal  to  the  rect- 
angle 1 AB  X  DG ; 
and  so  of  the  rest : 
therefore  the  sum  of 
the  rectangles  ^AB 
X  DG  +  |CB  X 
DF  +  |AC  X 
DE  is  equal  to  the 

whole  triangle  ABC  ;  but  the  sum  of  these  rectangles 
{by  Euc,  1.  2.)  is  equal  to  the  rectangle  under  half  the 
perimeter  AB  +  BC  +  AC  and  the  semi-diameter 
DG,  which  last  rectangle  is,  therefore,  equal  to  the 
triangle  given.  But  the  angles  E  and  G  being  right 
ones  {Eiic.  17.  3.),  and  the  side  AD  common,  and  also 
DE  equal  to  DG,  thence  will  AE  =  AG  {Euc.  47.  1.). 
And  in  the  same  manner  will  CE  =  CF  ;  consequently 
AC   CAE  +  CE)  will  be  =  AG   +  CF;     whence  it 

2Y 


346 


The  Construction  of 


appears  that  the  hypothenuse  is  less  than  the  sum  of  the 
two  legs  by  BG  -f  BF,  or  twice  the  radius  of  the  in- 
scribed circle,  and  therefore  less  than  half  the  perimeter 
by  once  that  radius,  or  DG :  whence  the  proposition  is 
manifest. 

PROBLEM  XXXV. 

The  perimeter  and  area  of  a  right-angled  triangle  being 
given^  to  describe  the  triangle. 

CONSTRUCTION. 

Make  AB  equal  to  the  given  perimeter,  which  bisect 
in  C,  and  upon  AC  let  the  rectangle  ACDE  be  consti- 
tuted equal  to  the  given  area  j  take  CF  =  CD,  and,  from 


Tu 


F,  through  D,  draw  the  indefinite  line  FH,  to  which,  from 
B,  apply  BI  =  AF  ;  then,  upon,  AB  let  fall  the  perpen- 
dicular IK,  so  shall  BIK  be  the  triangle  that  was  to  be 
constructed. 

DEMONSTRATION. 

Sinc^  {bij  construction^  CD  is  =  CF,  therefore  is  IK 
=  FK,  and  consequently  IK  +  IB  +  BK  =  FK  +  AF 
+  BK  =  AB.  Again,  the  excess  of  the  half  perimeter 
AC  above  the  hypothenuse  BI  ( AF)  being  =  CF  =  CD, 
it  is  evident  ^from  the  premised  lemma)  that  the  area  of 
the  triangle  will  be  =  ACDE  =  the  given  area  by  con- 
struction.    ^.  E.  D. 

Method  of  Calculation. 

Dividing  the  area  by  half  the  perimeter,  CD  (  =  CF) 
will  be  given  ;  then,  in  the  triangle  BFI,  will  be  given 
BF,  BI,  and  the  angle  F  (  =  45°) ;  whence  the  angle  B 
will  also  be  known,  and  from  thence  BK  and  BI. 


Geometrical  Problems. 


347 


PROBLEM  XXXVL 


To  make  a  right-angled  triangle  equal  to  a  given  square 
ABCD,  whose  sides  shall  be  in  arithmetical  progression. 

CONSTRUCTION. 

sAB 

In  AB,  produced,  take  BF  =  — — ,  and  upon  AF 

describe  the  semicircle    AEF,    cutting    BC,    produced, 


in  E  ;   take  BQ  = 
done. 


4EB 


;   join  E,  Q,  and  the  thing  is 


DEMONSTRATION. 

Since,  by  construction,  QB  :  BE  :  :  4  :  3,  therefore 
will  BQ2  :*EB2  :  :  16  :  9,  and  BQ^  +  BE*  :  BE^  :  :  16 
+  9  (25)  :  9,  that  is,  EQ^  :  BE^  :  :  25  :  9  {Euc.  47.  1.)  ; 
whence  EQ  ;  BE  :  :  5  .  3  {Euc.  22.  6.)  ;  therefore  the 
sides  BE,  BQ,  and  EQ,  being  to  one  another  in  the  ratio 
of  the  numbers  3,  4,  and  5,  are  in  arithmetical  progression. 


And,  because  BQ  is  = 
2EB2      2BF  X  AB 


4EB 


thence  will 


EB  x_BQ 
"2 


=  AB2.     ^  E.  D. 


3  3 

Method  of  Calculation. 

Seeing  BF  is  =  -±^,  (BE  V  AB  x  l^^)  will  be  = 


AB  V|;  whence  BQ  (i|^)  and  EQ  (^^ 
likewise  given. 

\ 


\  will 


be 


us 


Whe  Construction  of 


PROBLEM  XXXVII. 

In  a g'iveJi  circle  CIHK^  to  describe  three  equal  circles 
E,  F,  and  G,  ivhich  shall  touch  one  another^  and  also  the 
periphery  of  the  given  circle*  ^ 

CONSTRUCTION. 

From  the  center  C  let  the  right  lines  CH,  CI,  and 
CK  be  drawn,  dividing  the  periphery  into  three  equal 
parts,  in  the  points  H,  I,  and  K  ;  join  I,  K,  and  in 
CK,  produced,  take  KL  =  ^IK  j   draw  IL,  and,  paral- 


lel thereto,  draw  KF  meeting  CI  in  F  ;  make  HE  and 
KG  each  ==  IF,  and  upon  the  centers  F,  E,  and  G, 
through  the  points  I,  H,  and  K,  let   the    circles  FrI, 
EwH,  and  G/zK  be  described,  and  the  thing  is  done. 
DEMONSTRATION. 

Draw  FE,  FG,  and  EG. 

Because  {by  construction^  HE,  IF,  and  KG  are  equal, 
CE,  CF,  and  CG  will  likewise  be  equal,  and  FG  pa- 
rallel to  IK  {by  Euc.  2.  6.),  and  therefore,  KF  being 
parallel  to  IL  {by  construction^  the  triangles  IKL  and 
FGK  are  equiangular ;  whence,  IK  being  ==  2KL,  FG 
is  =  2GK  (2Fr)  {Euc.  4.  6.)  :  whence  it  is  manifest 
that  the  circles  F  and  G  touch  each  other. 


Geometrical  Problems.  349 

Moreover,  the  angles  ECF,  ECG,  and  FCG,  as  well 
as  the  containing  sides  CE,  CF,  and  CG  being  respectively- 
equal,  EF,  FG,  and  EG  must  also  be  equal  (Z>z/  Hue.  4. 1.), 
and  therefore  EF  or  EG  =  2FI  or  2GK ;  whence  it  is 
evident  that  the  circles  E,  F,  and  G  also  touch  one  ano- 
ther. But  all  these  circles  touch  the  given  circle,  because 
they  pass  through  given  points  H,  I,  K  in  its  periphery, 
and  have  their  centers  in  right  lines  joining  the  center  C 
and  the  points  of  concourse. 

Method  of  Calculation, 

In  the  triangle  FGK  we  have  given  the  angle  FGK 
(150°),  and  the  ratio  of  the  including  sides  (viz.  as  2 
to  1)  ;  whence  the  angle  FKG  will  be  given  ;  then,  in 
the  triangle  FCK  will  be  given  all  the  angles  and  the 
side  CK  ;  whence  CF  and  also  FI  will  be  given.  But, 
if  you  had  rather  have  a  general  theorem  for  expressing 
the  ratio  of  F I  to  CI,  then  let  EC  be  produced  to  meet 
FG  in  r.  Therefore  the  angle  rFC  being  =  30% 
Cr  will  be  =  |CF  ;  whence  {bi/  'Euc.  47.  1.)  FI  or 
Fr  (V  FC2  —_Cr^)  is  =  FC  X  VT,  and  therefore  CI 
=  FC  +  FC  V| ;  consequently  CI  :  FCj  :  1  -f^Vj:  1  ; 

whence,  by  division,  CI  :  FI  (  :  :  1  +  V  |   :  V~l)  •  •  '^  I 
+  1:1.  ^ 

PROBLEM  XXXVIII. 

In  a  given  circle  CEHG  to  describe  jive  equal  circles  K, 
L,  M,  N,  andOy  which  shall  touch  one  another^  and  the 
circle  given. 

CONSTRUCTION. 

Let  the  whole  periphery  EGH  be  divided  into  five 
equal  parts,  at  the  points  E,  F,  G,  H,  and  I  (by  Euc. 
11.2.),  and  draw  CE,  CF,  CG,  CH,  and  CI;  join  G, 
H,  and  in  CH  produced  take  HP  =  |.GH ;  draw  PG, 
and  parallel  thereto  draw  HM,  meeting  CG  in  M ; 
take  FL,  EK,  lO,  and  HN,  each  equal  to  MG,  and 
upon  the  centers  K,  L,  M,  N,  and  O,  let  circles  be 
described  through  the  points  E,  F,  G,  H,  and  I,  and  the 
thing  is  done. 


350 


The  Construction  of 


The  demonstration  whereof  is  evident  from  the  last 
proposition :  and  in  the  same  manner  may  6,  8,  or  10, 
£sPc.  equal  circles  be  inscribed  in  a  given  circle,  to  touch 
one  another. 

The  method  of  calculation  in  this,  or  in  any  other  case, 
will  also  be  the  same  as  in  the  last  problem  ;  for  in  the 
triangle  MNH  will  be  given  the  ratio  of  NM  to  NH 
(as  2  to  1),  and  the  included  angle  MNH  equal  to  126% 
120°,  11:3^°,  or  108°,  i^c.  according  as  the  number  of 
circles  is  5,  6,  8,  or  10,  ^c.  from  which  the  angle  MHN 
will  be  given;  then  in  the  triangle  CMH  will  be  given 
all  the  angles,  and  the  side  CH,  to  find  CM, 

PROBLEM  XXXrX. 

The  perimeter  of  a  right-angled  triangle^  whose  sides 
are  in  geometrical  progression^  being  given^  to  describe  the 
triangle. 

CONSTRUCTION. 

Upon  AC,  equal  to  the  giv..  n  perimeter,  describe  the 
semicircle  ABC,  and  let  AC  be  divided  in  D,  ac- 
cording to  extreme  and  mean  prcportion  ?  make  DB 
perpendicular   to    AC,    meeting    the    periphery   of  the 


Geometrical  Problems. 


351 


circle  in  B,  and,  having  joined  A,  B,  and  C,  B,  let 
AE  and  CE  be 
drawn  to  bisect 
the  angles  BAC, 
BCA ;  and,  from 
point  of  intersec- 
tion E,  let  EF  and 
EG  be  drawn  pa- 
rallel to  BA  and 
BC,  cutting  AC  in 
F  and  G  ;  then  will 
EFG  be  the  triangle  that  was  to  be  constructed. 

DEMONSTRATION. 

Since  (by  constructioii)  AC  :  AD  :  :  AD  :  DC ;  there- 
fore is  AC^  :  AC  X  AD  ::  AC  X  AD  :  AC  X  DC  {by 
Etic.  1.  6.),«or  ACy  :  ABq  :  :  ABq  :  BC<7  (^by  cor.  to  Euc. 
8.  6.),  and  consequently  AC  :  AB  :  :  AB  :  BC  ;  whence, 
the  triangles  ABC,  FEG,  being  equiangular,  FG  :  FE 
:  :  FE  :  EG.  Also,  EF  is  =  AF,  because  the  angle 
FEA  (=  EAB)  =  FAE  ;  and  in  the  very  same  man- 
ner is  EG  =  GC  ;  therefore  EF  +  FG  +  EG  (=  AF  + 
FG  +  GC)  =  AC.  Moreover,  the  angle  FEG  (=  ABC) 
is  a  right  angle,  by  Euc.  31.3.     ^  E.  D. 

Method  of  Calculation, 

Because  [by  construction)  AD  (=  x/^ACy  —  |AC)  =; 

AC  X  V4—  2.  thence  is  AB  ( \/ Ao  x  AD)  =  AC 

X   \f  VT  —  |.    and    BC    (VAC  x  CD  =  AD)  = 

AC  X  V I  —  I :  but,  by  reason  of  the  similar  tri- 
angles ABC,  FEG,  it  will  be,  as  AC  -f-  AB  +  BC 
•  (FG  +  FE  +  EG)  AC  :  :  AC  :  FG  :  :  AB  :   FE 

:  BC  :  EG ;  or  as  \f  v'J—  \  +  ^  +  Vj :  1   :  :  AC 
FG  :  :  AB  :  FE  :  ;  BC  :  EG ;  whence  FG,  FE,and  EG 
are  given. 


3j2 


The  Construction  of 


PROBLEM  XL. 

To  draw  a  right  line  PQ,  to  touch  two  circles  C  and  O, 
given  in  magnitude  and  position. 

CONSTRUCTION. 

Upon  the  line  CO,  joining  the  centers  of  the  given 
circles,  describe  the  semicircle  CDO,  in  which  inscribe 
CD  equal  to  the  difference  of  the  semi-diameters  CF  and 


OE  ;  and  from  the  point  B,  where  CD  produced  meets 
the  periphery  BF,  draw  PB  perpendicular  to  C/B  ;  then 
will  BP  touch  both  the  circles. 

DEMONSTRATION. 

Join  O,  D,  and  draw  O  A  perpendicular  to  PQ. 

The  angle  CDO,  standing  in  a  semicircle,  is  right ; 
therefore  the  angles  B  and  A  being  both  right  ones, 
by  construction,  the  angle  AOD  must  also  be  right, 
and  the  figure  DOAB  a  rectangle,  and  consequently 
AO  =  BD  =  BC—  CD  =  CF  —  CD  =  OE  {by  con- 
struction). Wherefore,  seeing  CB  and  OA  are  respec- 
tively equal  to  CF  and  OE,  and  both  the  angles  A  and  B 
right  ones,  it  is  evident  that  the  right  line  PQ  touches 
both  the  circles.     ^.  E.  D. 

The  numerical  solution  of  this  problem  is  extremely 
easy  ;  for  since  the  two  sides  CO  and  CD  of  the  right- 
angled  triangle  CDO  are  both  given,  the  angles  DCO  and 
AOC,  determining  the  points  of  contact  B  and  A,  are 
from  thence  given,  at  one  operation. 

But  if  it  be  required  to  draw  a  right  line  [ab)  to 
touch  both  circles,  and  to  pass  between  the  centers  C 


Geometrical  Problems*  353 

and  O ;  then,  instead  of  taking  CD  equal  to  the  differ- 
cnce  of  the  semi-diameters  CF,  OE,  let  Cd  be  taken 
equal  to  their  sum,  and  the  rest  of  the  process  will  be 
exactly  the  same, 

PROBLEM  XLL 

To  draw  a  right  line  AD  through  two  circles  GAEF, 
HCSR,  given  in  magnitude  and  position^  so  as  to  cut  off* 
segments  thereof^  AKBm,  CTDn,  equal  respectively  to  two 
given  segments  EQFa,  SPR^. 

CONSTRUCTION. 

Upon  the  subtenses  EF,  SR,  from  the  centers  G  and 
H,  let  fall  the  perpendiculars  GQ  and  HP  ;   and  from 


/72r 


the  same  centers,  at  the  distances  GQ,  HP,  let  two  cir- 
cles GQK,  HPT  be  described  ;  then  draw  a  right  line  AD 
to  touch  both  these  circles,  by  the  last  proposition,  and 
the  thing  is  done ;  for  the  lines  FE,  AB  being  at  the 
same  distance  from  the  center  G,  the  segments  cut  off  by 
them  must  consequently  be  equal :  and,  in  like  manner, 
the  segments  SPR6,  CTDw,  are  also  equaL 

PROBLEM  XLIL 

To  describe  the  circumference  of  a  circle  through  a  given 
point  P,  to  touch  two  right  lines  AB,  AC^  given  in  position. 

CONSTRUCTION. 

Join  A,  P,  and  bisect  the  angle  B AC,  with  the  right 
line  AK,  and,  from  any  point  Q  in  that  line,  draw  QT 

2Z 


354  The  Construction  of 

perpendicular  to  AC  ;    then,  from  Q  to  AP,  draw  QS 
=  QT  \  draw,  likewise,  PO  parallel  to  SQ,  meeting  AK 


in  O  ;    and  from  O,  as  a  center,  with  the  radius  OP,  de- 
scribe the  circle  PKF,  and  the  thing  is  done. 

DEMONSTRATION. 

Let  OH  be  perpendicular  to  AC,  and  OW  to  AB  ; 
then,  by  reason  of  the  parallel  lines,  it  will  be  QS  :  OP 
(:  :  AQ  :  AO)  :  :  QT  :  OH  ;  whence,  as  QT  =  QS^ 
OH  will  be  =  OP ;  and  therefore  the  circumference 
PKF  will  pass  through  the  point  H,  and  so,  AHO  be- 
ing a  right  angle,  AC  must  touch  the  circle  in  that 
point.  Moreover,  the  triangles  AOH  and  AOW  being 
equiangular,  and  having  one  side  common,  O  W  will  there- 
fore be  =  OH,  and  the  circle  also  touch  AB  in  the  point 
W.     ^.  E.  D. 

Mf^thod  of  Calculation, 

Having  assumed  AQ  at  pleasure,  there  will  be  given, 
in  the  triangle  AQT,  all  the  angles  and  one  side,  whence 
QT  (=  QS)  will  be  given:  then,  in  the  triangle  AQS, 
will  be  given  AQ,  QS,  and  the  angle  QAS,  whence  the 
angle  AQS  (=  AOP)  will  be  given.  Lastly,  in  the  tri- 
angle AOP  will  be  given  all  the  angles  and  the  side  AP, 
whence  AO  and  PO  will  be  given. 
Othertvise. 

Say>asthe  sine  of  OAH  :  radius  (:  ;  OH  :  OA  :  : 


Geometrical  Problems. 


^5$ 


OP  :  OA)  :  :  sine  of  OAP  :  sine  of  OP  A ;  then,  in 
the*  triangle  AOP  will  be  given  all  the  angles  and  the 
side  AP ;  whence  the  other  sides  AO  and  OP  will  be 
found. 

PROBLEM  XLIII. 

To  describe  the  circumference  of  a  circle  through  two  giv- 
en points  D,  G,  to  touch  a  right  line  AB,  given  in  position. 

CONSTRUCTION, 
Draw  DG,  and  bisect  the  same  by  the  perpendicular 
FC,  meeting  AB  in 
C  ;    join   C,   D,  and 
make    FP    perpendi- 
cular to    AB ;     and, 
from  F  to  CD,  pro- 
duced,   draw    FS  = 
FP;    make   DH  pa- 
rallel    to     FS,     and 
from    H,    the    inter- 
section   of    CF    and  j^  ^PT      C 
DH,  with  the  radius 
DH,  describe  the  circle  HDQ,  and  the  thing  is  done. 

DEMONSTRATION. 

Join  H,  G,  and  draw  HT  parallel  to  FP,  meeting  AB 
in  T  ;  then,  because  of  the  parallel  lines,  it  will  be,  FS  : 
HD  (:  :  CF  :  CH)  :  :  FP  :  HT ;  wherefore,  as  the  ante- 
cedents FS  and  FP  are  equal,  the  consequents  HD  and 
HT  must  likewise  be  equal ;  and,  therefore,  since  HT  is 
perpendicular  to  AB,  the  circumference  of  the  circle  will 
touch  AB  in  T  ;  and  it  will  also  pass  through  the  point  G, 
because  the  tw^o  triangles  DFH,  GFH,  having  two  sides 
and  the  included  angles  respectively  equal,  are  equal  in 
every  respect.   ^  E,  D. 

Method  of  Calculation. 

The  angle  FCA,  and  the  numbers  expressing  FC 
and  DG  being  given,  in  the  triangle  CFD  will  be 
given  (besides  the  right  angle)  both  the  legs  CF  and 
FD,  whence  CD  and  the  angle  FCD  will  be  known ; 


356 


The  Construction  of 


then  it  will  be,  as  the  sine  of  FC  A  (TCH) :  radius  (:  :  TH 
:  CH  :  :  DH  :  CH)  :  :  sine  of  HCD  :  the  sine  of  CDH  ; 
therefore  in  the  triangle  HCD  there  will  be  given  all  the 
angles  and  the  side  CD ;  whence  CH  and  HD  will  be 
known. 


PROBLEM  XLIV. 

Having  given  AB,  and  also  AD  and  ^G^  perpendicular 
to  AB  ;  X.0  find  a  point  T  in  AB,  to  which^  if  txuo  right 
lines  DT,  GT  be  drawn^  the  angle  DTG,  formed  by  those 
lines^  shall  be  the  greatest  possible* 

CONSTRUCTION. 

Describe,  by  the  last  problem,  a  circle  GDQ,  that  shall 
pass  through  G  and  D,  and  touch  AB,  and  the  point  of 
contact  T  will  be  the  point  required. 

DEMONSTRATION. 
Join  G,  T,  and  D,  T ;     and  from  any  other  point 

R,  in  the  line  AB,  draw  RG 
and  RD ;  also,  from  the 
point  Q,  where  GR  cuts  the 
circle,  draw  QD  ;  then,  the 
angle  GQD,  being  exter- 
nal with  regard  to  the  tri- 
angle DQR,  will  be  greater 
than  GRD  ;  therefore  GTD, 
standing  in  the  same  segment 
with  GQD,  will  be  also  great- 
er than  GRD.  ^  E.  D. 
Method  of  Calculation. 
DfRw  DE  parallel  to  AB ;  then  in  the  triangle  GDE 
will  be  given  DE,  EG  (  =  BG  —  AD),  and  the  right 
angle  DEG,  whence  the  other  angles  EDG,  EGD,  and 
the  side  DG  will  be  found;  then  in  the  triangle  CFP, 
similar  to  GDE,  will  be  given  all  the  angles  and  the  side 
FP  /=  AD4-BG\    ^^i^^j^^^  PC  ^iii  bg  gi^en;   from 

which,  by  proceeding  as  in  the  last  problem,  all  the  rest 
will  be  found. 


AK,      PT   CB 


Gesmetrical  Problems*  357 


PROBLEM  XLV. 


To  describe  a  circle^  which  shall  touch  two  right  lines 
AB,  AC,  g-iven  in  position^  and  also  another  circle  O, 
g'iven  in  magnitude  and  position* 

CONSTRUCTION. 

Let  the  angle  CAB,  made  by  the  concourse  of  the  two 
lines,  be  bisected  by  AK  ;  and,  from  any  point  P  in  this 
line,  let  fall  PQ  perpendicular  to,  AB,  which  produce  to 
R,  so  that  QR  may  be  equal  to  the  semi-diameter  of  the 


given  circle;  and  through  R,  parallel  to  AB,  draw  HM, 
meeting  KA  produced  in  H ;  draw  HO,  to  which,  from 
P,  draw  Pv  =  PR,  and  draw  OE  parallel  to  Fv^  meeting 
AK  in  E,  and  cutting  the  periphery  of  the  given  circle 
in  r ;  lastly,  from  E,  with  the  radius  Er,  describe  the 
circle  ErKN,  and  the  thing  is  done. 

DEMONSTRATION. 
Draw  EG  pei-pendicular  to  HM,  cutting  AB  in  F; 
then,  by  reason  of  the  parallel  lines,  PR  :  EG  (  :  :  HP 
:  HE)  :  :  Fv  :  EO ;  therefore,  PR  being  =  Pz;  {by 
construction)^  EG  and  EO  must  likewise  be  equal ;  from 
which  the  equal  quantities  FG  and  Or  being  taken 
away,  the  remainders  EF  and  Er  will  be  equal;    and 


358  "She  Construction  of 

therefore  the  circumference  rKN  passes  through  F  ;  but 
it  also  touches  AB  in  that  point,  because  EF  {by  construe- 
tioii)  is  perpendicular  to  AB  ;  it  likewise  touches  AC,  be- 
cause AE  bisects  the  angle  BAC  ;  lastly,  it  touches  the 
circle  O,  because  the  right  line  OE,  joining  the  centers 
O  and  E,  passes  through  the  point  r,  common  to  both  pe- 
ripheries. 

Method  of  Calculation 
Supposing  AO  drawn,  and  AS  perpendicular  to  HM, 
in  the  triangle  AHS  (besides  the  right  angle)  will  be 
given  AS  (=  rO)  and  the  angle  AHS  (=  EAF  = 
|BAC),  whence  AH  will  be  known;  then  in  the  tri- 
angle AHO  will  be  given  AH,  AO,  and  the  included 
angle,  whence  'AHO  and  HO  will  also  be  given :  then 
it  will  be,  as  the  sine  of  EHG  is  to  the  radius  ( ;  :  EG 
:  EH  :  :  EG  :  EH)  so  is  the  sine  of  EHG  to  the  sine 
of  EGH  ;  therefore  in  the  triangle  HEG  will  be  given  all 
the  angles  and  the  side  HO  ;  whence  EG  and  EH  are 
known  also. 

PROBLEM  XLVI. 

To  describe  the  circumference  of  a  circle  through  a  given 
point  P,  so  as  to  have  given  parts  cut  ojfby  two  right  lines 
AB,  hXL  given  In  position* 

CONSTRUCTION. 

Let  the  arcs  to  be  cut  off  by  AC  and  AB  be  similar 
respectively  to  the  arcs  ab^  be  of  any  given  circle  abcq^ 
whose  chords  ab^  be  subtend,  at  the  center,  any  given 
angles  aqb^  bqc.  Let  the  angle  ^Z>c  be  bisected  by  bdy 
take,  in  AB  and  AC,  any  two  pointy  E,  D,  equi- 
distant from  A  ,  and,  having  drawn  DE,  make  the  an- 
gle EDF  =  qbd,  CDR  =  qba,  and  BFR  =  qbc ;  then 
from  the  intersection  R  of  the  lines  DR  and  FR,  with 
the  radius  RD,  describe  an  arch  mS;z,  cutting  the  line 
AP  in  S,  draw  RS  and  ARK,  and  also  PQ,  parallel  to 
•  RS,  meeting  AK  in  Q  ;  then  from  the  center  Q,  with  the 
radius  PQ,  describe  the  circle  KPI,  and  the  thing  is 
done. 


Gejomctricdl  Problefns* 


359 


DEMONSTRATION. 

Draw  QH  and  QG  parallel  to  RF  and  RD,  meeting 
AB  and  AC  in  H  and  G.  The  angles  BED  and  CDE 
being  equal,  BFD  will  exceed  CDF  by  twice  EDF, 
or  by  twice  qbd^  that  is,  by  as  much  as  gbc  exceeds  qba^ 
or,  lastly,  by  as  much  as  BFR  exceeds  CDR  ;  therefore, 
seeing  the  whole  angle  BFD  as  much  exceeds  the  whole 
angle  CDF,  as  the  part  BFR  of  the  former  exceeds  the 
part  CDR  of  the  latter,  the  remaining  parts  RFD  and 
RDF  must  be  equal,  and  consequently  FR  =  RD  =  RS» 
But,  by  reason  of  the  parallel  lines,  it  will  be,  RF  :  QH  :  : 
RD  :  QG  :  :  RS  :  QP ;  whence  the  antecedents  RF, 
RD,  RS,  being  equal,  the  consequents  QH,  QG, 
QP,  must  be  also  equal,  and  the  circumference  must 
pass  through  the  points  H  and  G ;  whence  the  solution 
is  manifest. 

Method  of  Calculation. 

If  two  perpendiculars  be  conceived  to  fall  from  Q 
upon  AB  and  AC,  they  will,  it  is  plain,  be  in  the  given 
ratio  of  the  sines  of  the  angles  QHI  and  QGL ;  there- 
fore the  position  of  the  line  AQK  will  be  given  {fro?n 


360 


The  Construction  of 


prob.  26)  by  saying,  as  the  sum  of  the  said  sines  is  to  their 
difference,  so  is  the  tangent  of  half  BAC  to  the  tangent  of 
halfBAQ— CAQ. 

Again,  it  will  be  as  sine  QAH  :  sine  QH A  (:  :  QH  : 
QA  :  :  QP  :  QA)  :  :  sine  QAP  :  sine  QPA  ;  therefore, 
in  the  triangle  AQP,  are  given  all  the  angles  and  one  side 
AP,  whence  AQ  and  PQ  will  be  found. 

PROBLEM  XLVIL 

Having  the  three  perpendiculars^  let  fall  from  the  angles 
of  a  plane  triangle  on  the  opposite  sides  ^  equal  to  three  given 
right  lines  K^,  L/,  and  Mw,  to  describe  the  triangle* 

CONSTRUCTION. 
Draw   the  indefinite    right    line    RS,    in    which  take 
AB  equal  to  O :     find  a  fourth  proportional  to   M;w, 
L/,  and  K^,  with  which,  as  a  radius,  from  the  center  A, 


let  an  arch  rCs  be  described;  and  from  B,  with  "the  ra- 
dius L/,  let  another  arch  be  described  intersecting  the  for- 
mer in  C  ;  join  A,  C,  and  B,  C,  and  upon  RS  let  fall  the 
perpendicular  QC,  in  which,  produced,  take  QP  =  L/, 
and  draw  PF  parallel  to  RS,  meeting  AC,  produced,^  in 
F,  draw  FG  parallel  to  CB,  and  AFG  will  be  the  triangle 
required. 

DEMONSTRATION. 
Draw  FE,  Gjr,  and  Av  perpendicular  to  the  three  sides 
©f  the  triangle. 


Geometrical  Problems. 


361 


The  triangles  ABC,  AGF ;  AFE,  AG^ ;  and  GFE, 
AGz;,  are  equiangular,  by  construction;    therefore  Gq  : 

FE  :  :  AG  :  AF  :  :  AB  (K>^)  :  AC  i^-^)  -  •  Mm 

:  L/ ;  whence,  as  the  consequents  FE  and  hi  are  equal, 
by  construction^  the  antecedents  Gq  and  Mm  must  be 
equal  likewise.  Again,  BC  (L/)  :  AB  (Ky^j  (:  :  FG 
:  AG)  :  :  FE  (L/)  :  A^ ;  and  consequently  Kk  =  Kv. 
4  E.  D. 

Method  of  Calculation. 

Since  K/f,  L/,  and  Mm  afe  given,  AC  f  =  — =^-- — I 

will  be  known ;  then  in  the  triangle  ABC  will  be  given 
all  the  three  sides,  whence  the  angles  are  known  ;  lastly, 
in  the  triangle  AFG  will  be  given  all  the  angles  and  the 
perpendicular  EF,  whence  the  sides  are  also  known. 


PROBLEM  XLVIII. 

The  position  of  three  points^  in  the  same  right  line^  being 
given^  it  is  proposed  to  find  a  fourth^  where  lines^  drawn 
from  the  former  three^  shall  make  given  angles  with  each 
other. 

CONSTRUCTION. 

Let  the  three  given  points  be  A,  B,  and  C  :  make 
the  angles  ACE  and 
C  AE  respectively  equal 
to  the  given  angles 
which  the  lines  drawn 
from  B,  A,  and  B,  C 
are  to  make ;  and  let 
AE  and  CE  meet  in 
E  ;  through  A,  C,  and 
E,  let  the  circumference 
of  a  circle  AE'CD  be 
described,  and,  through 
E  and  B,  draw  EBD, 
meeting  it  in  D,  then 
will  D  be  the  point  re- 
quired. 

3  A 


36^ 


The  Construction  of 


DEMONSTRATION. 
Join  A,  D,  and  C,  D. 

The  angle  EDA  is  equal  to  ACE,  standing  on  the 
same  segment  j  and  for  the  like  reason  is  EDC  =  CAE. 
4  E.  D. 

Method  of  Calculation. 

In  the  triangle  ACE  are  given  all  the  angles  and  the 
side  AC,  whence  AE  will  be  given ;  then  in  the  triangle 
ABE  will  be  given  the  two  sides  AE,  AB,  and  the  in- 
cluded angle  ;  whence  ABE  and  all  the  rest  of  the  angles 
in  the  figure  will  be  given. 


D  A, 


PROBLEM  XLIX. 

Three  points^  A,  B,  C,  being  any  how  given  ;  to  find  a 
fourth^  where  lines^  drawn  from  the  former  three^  shall 
make  given  angles  with  one  another. 

CONSTRUCTION. 

Join  the  given  points, 
and  upon  the  right  line 
AB  describe  a  segment  of 
a  circle,  capable  of  the 
given  angle  which  that 
line  is  to  subtend ;  com- 
plete the  circle,  produce 
BA,  and  make  the  angle 
DAQ  equal  to  the  angle 
which  BC  is  to  subtend, 
and  let  AQ  meet  the  pe- 
riphery in  Q ;  draw  QC, 
cutting  the  same  periphery 
in  P  ;  join  A,  P,  and  B, 
P,  and  the  thing  is  done. 

DEMONSTRATION. 

The  angle  ABP  is  equal  to  the  given  angle  which  AB 
was  to  subtend  {by  construction)  ;  and  the  angles  QAB 
and  QP  A,  standing  upon  the  same  segment,  being  equal  to 
each  other,  their  supplements  DAQ  and  BPC  must  like- 
wise be  equal,     i^.  E.  D. 


Geometrical  Problems* 


363 


Method  of  Calculation. 
Join  B,  Q ;  then,  in  the  triangle  ABQ  will  be  given  all 
the  angles  and  the  side  AB,  whence  BQ  and  ABQ  will  be 
known  ;  then  in  the  triangle  CBQ  will  be  given  two  sides, 
and  the  included  angle  CBQ;  whence  the  angle  CQB, 
equal  to  BAP,  will  be  known  ;  lastly,  in  the  triangle  APB 
will  be  given  all  the  angles  and  the  side  AB,  from  which 
AP  and  BP  will  be  found. 


PROBLEM  L. 

To  draw  a  right  line  EG  through  a  circle  O,  given  in 
magnitude  and  position^  which  shall  also  cut  a  right  line 
Q^C^  given  in  position^  in  a  given  angle^  and  have  its  parts 
EF,  FG  intercepted  hy  the  circle  and  that  right  line^  in  the 
given  ratio  of  the  two  right  lines  ah  and  he. 

CONSTRUCTION. 

At  any  point  B,  in  the  right  line  QC,  make  the  angle 
QBA  equal  to  the  given  angle,  and  through  the  center 
O,  perpendicular 
to  BA,  draw  DQ 
meeting  BA  in 
R,  and  CG  in 
Q ;  bisect  ah  in 
d^  and  in  RB  take 
R^  =  bd^  and  pq 
=  ^c,  and  draw 
pm  and  qn  paral- 
lel to  DQ ;  from 
the  point  72,  v/here 
qn  intersects  QC, 
draw  TzL  parallel 
to  BA,  meeting 
pm  in  m ;  through 
the  points  Q  and 
m  draw  QmF, 
cutting  the  periphery  of  the  circle  in  F,  and  through  F,  pa- 
rallel to  BA,  draw  EFG,  and  the  thing  is  done. 
DEMONSTRATION. 

The  lines  GE,  BA,  and  nL,  being  parallel,  the  an- 


364  The  Construction  of 

gles  QGE,  QB  A,  &>V.  will  be  equal,  and  likewise  SF  : 
FG  :  :  Lm  :  mn ;  but  Lw  {by  construction')  is  (  =  "Rp) 
=  dh^  and  mn  (=pq)  ==  be ;  therefore  SF  :  FG  :  i  db  : 
be,  and  consequently  EF  (2SF)  :  FG  :  :  ab  (2bd)  :  be. 

Method  of  Calculation* 
L,7i  {dc^  :  Lw  {db)  :  :  the  tangent  of  LQ/z  (the  comple- 
ment of  the  given  angle  QBR)  :  the  tangent  of  LQw ; 
therefore,  in  the  triangle  OQF  will  be  given  one  angle 
OQF  and  two  sides,  QO,  FO  ;  whence,  not  only  the 
angle  SOF,  but  also  SO  and  SF  will  be  known. 

PROBLEM  LI. 

To  apply,  or  inscribe^  a  given  right  line  AD  betrveen  the 
peripheries  of  two  circles  C  and  O,  given  in  magnitude  and 
position,  so  as  to  be  inclined  to  the  right  line  CO,  joining  the 
centers,  in  a  given  angle, 

CONSTRUCTION. 

Make  OCB  equal  to  the  given  angle,  and  let  CB  be 
taken  equal  to  the  given  line  ;  upon  the  center  B,  with 
the  radius  of  the  circle  C,  let  the  arch  nDm  be  describedy 


^J>^ 


cutting  the  circle  O  in  D  :  then  draw  BD,  and,  parallel 
thereto,  draw  CA,  meeting  the  periphery  in  A  ;  join  A, 
D,  and  the  thing  is  done. 

DEMONSTRATION. 
Because  {by  construction^  CA  and  BD  are  equal  and 
parallel,  there iore  will  AD  and  CB  be  also  equal  and  pa- 
rallel {by  £uc.  33.  1.).     ^  £.  D. 


Geometrical  Problems'^ 


365 


Method  of  Calculation* 
In  the  triangle  CBO  are  given  two  sides,  CO  'and  CB, 
and  the  angle  OCB  ;  whence  OB  and  the  angle  COB  will 
be  known  ;  then  in  the  triangle  OBD  will  be  given  all  the 
three  sides,  whence  all  the  angles,  and  consequently  DOC^ 
will  also  be  known. 

PROBLEM  LIL 

From  a  given  rectangle  ABCD,  to  cut  off  a  gnomon 
ECG,  zvhose  breadth  shall  be  every  where  the  same^  and 
whose  area  shall  be  just  half  that  of  the  rectangle* 

CONSTRUCTION. 
In  BA  take  BH  equal  to  BC,  or  AD ;  and  in 
DA,  produced,  take 
AP  a  mean  propor- 
tional between  B  A  and 
|AD  (so  that  AP2 
may  =  the  given  area 
AGFE).  From  P  to 
the  middle  of  AH 
draw  PO  ;  make  OE 
=  OP,  and  DG  = 
BE ;  complete  the 
rectangle  EAGF,  and 
the  thing  is  done. 

DEMONSTRATION. 

If  the  semicircle  EPQ,  from  the  center  0,be  described, 
it  is  plain  that  AQ  =  EH  =  BH  —  BE  =  AD  —  DG  = 
AG ;  and  consequently  that  AE  x  AG  =  AE  X  AQ  = 
AP2  {Euc.  13.  6.)     ^  E.  D. 


(= 


Method  of  Calculation, 

In    the    right-angled    triangle    AOP  are  eriven    AO 
AB  — BC\ 


-j  and  AP  (=  V^AB  x  BC) ;    whence 


OP  will  be  known,  and  from  thence  both  AE  and  AG. 


366  The  Construction  of 

PROBLEM  LIIL 

Three  points^  A,  B,  C,  being  given  j  it  is  proposed  to  find 
a  fourth^  Y^from  xvhence  lines^  drawn  to  the  three  former^ 
shall  obtain  the  ratio  of  three  given  lines  a,  b^  and  c^  respec- 
tively. 

CONSTRUCTION. 
Having  joined   the  given  points,    take   AF,  in  AB, 
equal  to  «,  and  AI  =  Cj    also  make  the  angles   AFG 

C^  and   A  IK   equal, 

each,  to  ACB ; 
and  from  the  cen- 
ters F  and  G, 
with  the  radii  b 
and  AK  respec- 
tively, let  two 
arcs  be  described 
intersecting  in  H; 
from  which  point 
draw  HF  and 
HA  ;  then  draw  BP  to  make  the  angle  ABP  =  AHF,  and 
it  will  meet  AH  (produced)  in  the  point  P,  required. 
DEMONSTRATION. 
Let  BP,  CP,  and  GH  be  drawn.  The  triangles 
ABP,  AHF  being  equiangular  {by  constructio7i\  it  will 
be  AP  :  BP  :  :  AF  (a)  :  FH  (b) ;  also  AB  :  AP  :  :  AH 
:  AF  ;  and  AB  :  AC  :  :  AG  :  AF  (because  ABC  and 
AGF  are  likewise  equiangular)  ;  whence  it  is  evident, 
since  the  extremes  of  the  two  last  proportions  are  the 
same,  that  AP  x  AH  =  AC  X  AG,  or  AC  :  AP  :  :  AH 
:  AG  ;  therefore  the  triangles  ACP,  AHG  being  equian- 
gular (Euc.  6.  6.),  we  have  AP  :  CP  :  :  AG  :  GH  (AK) 
:  :  AF  (a)  :  AI  (c).     ^  E.  D. 

Method  of  Calculation. 
In  the  triangles  AFG,  AIK  are  given  all  the  angles 
and  the  sides  AF  and  AI,  whence  AG,  FG,  and  AK 
(GH)  will  be  found ;  then  in  the  triangle  FGH  will  be 
given  all  the  sides,  to  find  the  angle  HFG  ;  which,  added 
to  AFG,  gives  AFH  (APB),  from  whence,  and  the  two 
given  sides  AF  and  FH  including  it,  every  thing  else  is 
readily  determined. 


Geometrical  Problems.  367 


PROBLEM  LIV. 

To  describe  a  triangle  (ABC)  similar  to  a  given  one 
AMN,  such  that  three  lines  (AP,  BP,  CP)  may  be  drawn 
from  its  angidar  points  to  meet  the  same  point  (Pj,  so  as  to 
be  equal  to  three  given  lines  AD,  AF,  and  AK,  respec- 

tivelij* 

CONSTRUCTION. 

Draw  DE  and  KG,  making  the  angles  ADE 
and  AKG,  each,  equal  to  the  given  angle  N,  and 
intersecting  AN  in 
E  and  G  ;  from  the 
centers  D  and  E, 
with  the  intervals 
AF  and  AG,  let  two 
arcs  be  described, 
intersecting  in  H ; 
draw  AH,  in  which 
take  AP  =  AD; 
andfrom.P,  to  AM 
and  AN,  apply  PB  A. 
and  PC  equal,  respectively,  to  AF  and  AK,  and  let  B,  C, 
be  joined  ;  so  shall  ABC  be  the  triangle  that  was  to  be  de- 
termined, 

DEMONSTRATION. 

The  three  lines  AP,  BP,  CP  are,  respectively,  equal 
to  the  three  given  lines  AD,  AF,  AK,  by  con^aiction ; 
we  therefore  have  only  to  prove  that  the  triangle  ABC  is 
similar  to  the  given  one  AMN.  Now,  supposing  DH  and 
EH  to  be  drawn,  it  will  be  AP  :  PC  (or  AD  :  AK)  :  : 
AE  :  AG  (EH)  ;  whence  the  triangles  APC  and  AHE 
will  be  equiangular  {Euc*  6.  6.),  and  consequently  AC  : 
AH  :  :  AP  (AD)  :  AE  :  :  AN  :  AM  {Euc.  5.  6.) ;  but 
the  triangles  ABP  and  ADH  (having  AP  =  AD,  PB  = 
DH  (^2/  construction)^  and  the  angle  DAP  common)  are 
equal  in  all  respects  ;  therefore,  by  substituting  AB  in  the 
room  of  AH,  our  last  proportion  becomes  AC  :  AB  :  ; 
AN  :  AM  ;  whence  it  is  manifest  that  the  triangles  ABC 
and  AMN  are  equiangular.     ^.  E.  D. 


368  The  Construction  of 

Method  of  Calculation* 
In  the  triangles  ADE,  AKG  are  given  all  the  an- 
gles and  the  sides  AD  and  AK,  from  which  AE,  DE» 
and  AG  will  be  known ;  then  in  the  triangle  DHE 
will  be  given  all  the  sides,  to  find  the  angle  EDH> 
which  added  to  ADE  gives  ADH  ;  from  whence,  and 
the  two  given  sides  including  it,  AH  (=  AB)  will  be 
known. 

PROBLEM  LV. 

In  the  triangle  ace^  besides  the  angle  c,  are  given  the  seg- 
ments  of  the  sides  ab  and  de^  and  the  angles  aeb  anddbe  sub- 
tended thereby;  to  describe  the  triangle* 
CONSTRUCTION. 

Upon  AB,  equal  to  ab^  let  a  segment  of  a  circle  be 
described  to  contain  an  angle  equal  to  aeb :  make  the 
angle  ABF  =  ace^  BA?2  =  dbe^  and  the  line  BF  =  ^f/; 


from  the  point  n^  where  A/z  cuts  the  periphery  of  the 
circle,  through  F,  draw  nFE,  meeting  the  periphery  in 
E  ;  join  A,  E,  and  B,  E,  and  draw  EC  parallel  to  BF, 
meeting  AB,  produced,  in  C  ;  and  then  the  thing  is 
done. 

DEMONSTRATION. 

Let  BD  be  parallel  to  FE. 

Since  the  lines  BD,  EF,  and  ED,  FB  are  parallel, 
therefore  is  ED  =  BF  (  =  ed)^  and  the  angle  ACE  also 
=  ABF  {ace)  £uc.  28.  1.      Moreover,  the  angle  BEn 


Geometrical  Problems.  369 

(DBE)  is  equal  to  BAn  (dbe\  both  standing  upon  the 
same  segment  Bn.     ^.  E.  I). 

Method  of  Calculation. 
Join  B,  n ;  then  in  the  triangle  AB^i  will  be  given  all 
the  angles  and  the  side  AB,  whence  B/z  will  be  known ; 
then  in  the  triangle  nBF  will  be  given  B/z,  BF,  and  the 
included  angle  /zBF,  whence  BF;2  (CDB)  and  all  the  rest 
of  the  angles  in  the  figure  will  be  known. 

PROBLEM  LVI. 

To  make  a  trapezium^  whose  diagonals^  and  two  oppO' 
site  sides ^  shall  be  all  of  given  lengths^  and  whereof  the  an- 
gle formed  by  the  given  sides ^  when  produced  till  they  meet^ 
shall  also  be  given. 

CONSTRUCTION, 
Draw  the  indefinite  right  line  AC,  and  take  therein 
AB  equal  to  one  of  the  two  given  sides  ;  make  the  angle 
CBG  equal  to 

the  given  an-      q^  C\ 

gle,andletBG      t"^ 
be  made  equal       \ 
to    the    other        \ 
given  side;  up-         *5 
on  the  centers 
A  and  G,  with 
intervals  equal 
to  the  two  dia- 
gonals, let  two 
arches  be  de- 
scribed,    cut- 
ting each  other 

in  D  ;  make  DE  equal  and  parallel  to  GB  ;  join  D,  B, 
and  E,  A  :  then  ABDE  will  be  the  trapezium  required. 

DEMONSTRATION. 
Draw  DG,  DA,  and  BE,  and  let  BA  and  DE  be  pro- 
duced to  meet  each  other  in  F. 

The  lines  BG  and  DE  are  equal  and  parallel,  by- 
construction  J    therefore  BE  is  =  DG,  which  last  (by 

3  B 


^fO  The  Construction  of 

construction)  is  equal  to  one  of  the  given  diagofials,  as 
AD  is  equal  to  the  other  ;  moreover,  the  sides  AB  and 
ED  (BG)  are  equal  to  the  given  sides,  by  construction ; 
and  the  angle  F  is  equal  to  the  given  angle  CBG,  because 
DF  is  parallel  to  GB.     ^  E.  D. 

Method  of  Calculation, 
Suppose  AG  to  be  drawn  ;  then  in  the  triangle  ABG 
will  be  given  the  two  sides  BA  and  BG,  and  the  included 
angle  ABG ;  whence  the  side  AG  and  the  other  two 
angles  will  be  known ;  then  in  the  triangle  ADG  will  be 
given  all  the  sides ;  whence  the  angle  AGD  will  be  known, 
and  from  thence  the  whole  angle  BGD  ;  lastly,  in  the 
triangle  BGD  wuU  be  given  the  two  sides  BG  and  GD, 
and  the  included  angle  BGD ;  whence  the  side  BD  wiU 
likewise  be  known. 


PROBLEM  LVIL 

The  segments  of  the  base  AD,  DB,  and  the  line  DC, 
bisecting  the  vertical  angle  ACB  of  a  plane  triangle^  being 
given^  to  describe  the  triangle* 

CONSTRUCTION. 
In  AB,  produced,  take  DO  to  AD,  as  DB  to  AD 

--^  r  —    ^^'    ^^^ 

from  the  cen- 
ter O,  with  the 
radius  OD,  de- 
scribe the  cir- 
cle DCQ ;  also 
from  the  center 
D,  at  the  given 
distance     DC, 
describe  the  circle   mCn^  and  from   C,  the   intersection 
of  the  two  circles,  draw  CA  and  CB,  and  the  thing  is 
done. 

DEMONSTRATION. 
Since  DO  :  AD  :  :  DB  :  AD  —  DB  ;    therefore  (bij 
the  lemma  in  p.  334.)  AC  :  CB  :  :  AD  :  DB  ;  whence  CD 
bisects  the  angle  ACB  {bij  Euc.  3.  6. j.     ^.  E.  D. 


Geometrical  Problems, 


371 


Method  of  Calculation^ 
Draw  CP  perpendicular  to  AQ. 

Because,   by    construction,    OD    is  =  -r-— tt^; 

'     ^  AD  —  BD 

Aerefore  will  DQ  =  ^^  ;    whence,  by  reason 

AU  — —  JljJL) 

of   the  similar    triangles    DCQ,    DPC,    it    will  be,    as 


2ADx  BD 


DC 


AD  —  BD 

whence  AC  and  CB  are  given 


DC : DP  = 


AD  —  BD  X  DC^ 

2AD  X  BD 


PROBLEM  LVIIL 

Having  given  the  base^  the  angle  at  the  vertex^  and  the 
Une  drawn  from  thence  to  bisect  the  base  ;  to  construct  the 
•triangle. 

CONSTRUCTION. 

Upon  the  given  base  AB 
describe  {by  prob.  4.)  a  seg- 
ment of  a  circle  ADB  ca- 
pable of  the  given  angle ; 
and,  from  the  point  F,  in 
which  the  perpendicular 
DF  bisects  AB,  with  a  ra- 
dius FC  equal  to  the  bisect- 
ing line,  describe  nCm^  cut- 
ting the  periphery  ACB  in 
C  ;  join  A,  C,  and  B,  C,  and 
the  thmg  is  done. 

The     demonstration     of 
which  is  evident  from  the  construction. 

Method  of  Calculation, 
From  the  center  O  let  OA  and  OC  be  drawn;  then 
in  the  triangle  AOF  will  be  given  all  the  angles  and 
the  side  AF  ;  whence  FO  and  OC  (OA)  will  be  known, 
and  in  the  triangle  CFO  will  be  given  all  the  sides  ; 
whence  the  angle  FOC,  and  its  supplement  DOC,  ex- 
pressing the  difference  of  the  angles  at  the  base,  will  also 
be  known. 


372 


The  Construttion  of 


PROBLEM  LIX. 

The  base  J  the  difference  of  the  angles  at  the  base,  and  the 
line  drawn  from  the  vertical  angle  to  bisect  the  base  of  any 
plane  triangle^  being  given  ;  to  describe  the  triangle* 

CONSTRUCTION. 

Upon  AB,  equal  to  the  given  base,  let  a  segment  of  a 
circle  AHEB  be  described  to  contain  an  angle  equal  to 

the  difference  of  the 
JS.  ^\  angles  at  the  base ; 
bisect  AB  in  C,  and 
take  CD  to  AC  in  the 
duplicate  ratio  of  AC 
to  the  given  bisecting 
line  KL ;  make  CS. 
and  DI  perpendicular 
to  AB,  cutting  the 
circle  in  S  and  I ;  draw 
A  I,  cutting  CS  in  G  ; 
and  through  G  draw 
the  chord  EGH  paral- 
lel to  AB  ;  join  A,  E,  and  A,  H,  and  in  AI  take  AN 
equal  to  KL;  draw  MNP  parallel  to  EH,  meeting  AE 
and  AH  in  M  and  P  ;  then  will  AMP  be  the  triangle 
which  was  to  be  constructed. 

DEMONSTRATION. 

Since  (by  construction)  CG  is  parallel  to  DI,  and  KL^ 
:  AC^  :  \  AC  :  CD  ;  therefore  {Eiic.  4.  6.)  KL^  : 
ACq  :  :  AG  :  GI  :  :  AGy  :  GI  X  AG ;  but  GI  x  AG 
=  EG  X  GH  =  EG^  {Euc.  35.  3.  and  3.  3.)  ;  therefore 
KL^  :  AC^  :  ;  AG^  :  EG^  ;  and  consequently  KL  : 
AC  :  :  AG  :  EG  :  :  AN  :  NM  ;  but  AN  is  {by  con- 
struction)  equal  to  KL,  therefore  NM  is  =  AC,  and 
consequently  MP  (2MN)  =  AB.  Moreover,  the  dif- 
ference of  the  angles  at  the  base,  P  —  M,  is  (  =  AHE  — 
AEH)  =  AEB  ;  which  {by  construction^  is  equal  to  the 
difference  given.     ^.  E.  D, 

Method  of  Calculation. 

From  the  center  O  draw  OA  and  OI ;  also  draw  li> 
parallel  to  EH,  meeting  OS  in  v :    then  it  will  be  {by 


Geometrical  Problems* 


373 


Construction)  as  KLy  :  AC^  ( :  :  AC  :  Iv)  :  :  the  sine 
of  AOC  or  AEB,  the  given  difference  of  the  angles  at 
the  base,  to  the  sine  of  SOI,  which,  added  to  AOS,  gives 
AOI,  whose  supplement,  divided  by  2,  will  be  OIG ; 
from  whence  OGI  and  its  supplement  OGA  are  given ; 
and  consequendy  ANM  (equal  to  AGE)  ;  then  in  the 
triangle  ANM  will  be  given  AN,  NM,  and  the  includ- 
ed angle  ANM,  whence  the  angles  M,  A,  P  will  also 
be  given. 

PROBLEM  LX. 


The  perpendicular^  the  angle  at  the  vertex^  and  the  sum 
of  the  three  sides  of  a  triangle  being  given  ;  to  describe  the 
triangle. 

CONSTRUCTION. 

Make  AB  equal  to  the  sum  of  the  sides,  which  bi- 
sect in  P,  making  PO  perpendicular  to  AB,  and  the 
angle  PAO  equal  to  half  the  given  angle  at  the  ver- 
tex ;  from  the  center 
O,  with  the  radius 
OA  describe  the  cir- 
cle AHB,andinOP, 
produced,  take  PK 
equal  to  the  given  per- 
pendicular, and  draw 
KH  parallel  to  BA, 
cutting  the  circle  in  H; 
join  A,  H,  and  B,  H, 
and  make  the  angles 
BHF  and  AHE  equal  to  HBF  and  HAE,  respectively ; 
then  will  EHF  be  the  triangle  required. 

DEMONSTRATION. 

Join  O,  B,  and  O,  H,  and  draw  HQ  perpendicular  to 
AB. 

The  angle  EFH  is  =  BHF  +  HBF  ==  2HBF  {by 

construction)  =  HOA  {Euc.  20.  3.)  :  and,  in  the  same 
manner,  is  FEH  =  HOB  ;  hence  it  follows  that  EFH 
+  FEH  (=  HOA  +  HOB)  =  AOB ;  and,  by  taking 
each  of  these    equal  quantities  from  two  right  angles, 


374 


The  Construction  of 


we  have  EHF  =  OAB  +  OBA  {Euc.  32.  1.)  =  20AB 

^  ==  the  given  angle  {by  construction^ •  Moreover^  QH  is 
=  PK  —  the  given  perpendicular;  and,  EH  being  = 
AE,  and  FH  =  BF  {Euc.  6-  !.)>  EH  +  HF  +  EF 
will  therefore  be  =  AB  =  the  given  sum  of  the  sides* 

^  ^-  ^-  . 

Method  of  Calculation. 

In  the  triangle  AOP  are  given  all  the  angles  and  the 
side  AP,  whence  OP  and  AG  (HO)  are  known  :  then  in 
the  triangle  OHK  will  be  given  the  sides  OH  and  OK 
(OP  +  PK),  whence  HK  w^ill  be  given  ;  next,  in  the 
triangle  BQH  will  be  given  QH  and  BQ  (BP  —  HK), 
whence  QBH,  and  its  double  QFH,  w^ill  be  given  ;  lastly, 
in  the  triangle  EFH  are  given  all  the  angles  and  the  per- 
pendicular Qi-i,  w^hen'ce  the  sides  will  also  be  given. 

But  the  answer  may  be  more  easily  brought  out,  by 
first  finding  HOK,  the  difference  of  the  angles  ABH  and 
BAH,  as  in  the  fifth  problem. 

PROBLEM  LXI. 


The  sum  of  the  three  sides^  the  difference  of  the  angles 
at  the  base^  and  the  length  of  the  line  bisecting  the  vertical 
angle  of  any  plane  triangle  being  given  ;  to  describe  the  tri- 
angle. 

CONSTRUCTION. 
Make  AB  equal  to  the  sum  of  the  sides,  wliich  bisect 
in  E  by  the  perpendicular  DE^z,  and  make  the  angle  yzE?" 

equal  to  half  the 
given  difference 
of  the  angles  at 
the  base,  taking 
Er  equal  to  the 
line  bisecting  the 
vertical  angle  : 
through  r  draw 
Cnr  parallel  to 
AB,  cutting  DE7Z 
in  n ;  draw  tzA, 
to  which  draw  Em  =  Er,  and  draw  AD  parallel  to  Ew, 
meeting  nED  in  D  ;  and  on  the  center  D,  at  the  distance 


Geometrical  Problems*  375 

of  DA,  describe  the  circle  ACB,  Cutting  rnC  in  C  ;  join 
A,  C,  and  B,  C,  and  make  the  angle  BCF  =  CBF  ;  also 
make  ACG  =  CAG,  and  let  CF  and  CO  meet  AB  in 
F  and  G  ;  then  will  FCG  be  the  triangle  that  was  to  be 
described. 

DEMONSTRATION. 

Upon  AB  let  fall  the  perpendicular  CP  ;  let  CQ  bisect 
the  vertical  angle  GCF,  and  let  DH  be  drawn  parallel 
to  Er,  meeting  Cr  in  H.  Then,  by  reason  of  the  paral- 
lel lines,  it  will  be  as  Er  :  DH  (:  :  En  :  D;i)  :  :  Ew  :  DA  5 
whence,  Er  being  =  Ew  (by  constructwii)^  DH  and  DA 
are  also  equal,  and  the  point  H  falls  in  the  periphery  of 
the  circle  :  therefore  the  angle  ;?DH  (nEr)  at  the  cen- 
ter, standing  upon  half  the  arch  HC,  will  be  equal  to  the 
angle  HAC,  at  the  periphery,  standing  upon  that  whole 
arch,  that  is,  equal  to  the  difference  of  the  angles  ABC 
and  BAG  ;  but  the  angle  GFC  being  double  to  ABC, 
and  FGC  double  to  BAG  {by  construction)^  the  differ^ 
ence  of  GFC  and  FGC  will  be  double  to  the  difference 
between  ABC  and  BAG,  and  therefore  equal  to  2;zEr 
(2?zDH},  the  difference  given.  Moreover,  because  GCQ 
=  FCQ,  2PCQ  will  be  the  difference  between  PCG  and 
PCF,  which  must  likewise  be  equal  to  2;2Er,  the  differ- 
ence of  their  complements  PGC  and  PFC  ;  whence  PCQ 
=  ;zEr,  and  consequently  CQ  =  Er.  Furthermore,  since 
the  angle  ACG  =  CAG,  and  BCF  =  CBF,  thence  will 
CG  =  AG,  and  CF  =  FB  5  and  therefore  CG  +  GF  + 
FC  =  AB.     ^  E.  D. 

Method  of  Calculation. 
In  the  triangle  E/zr  are  given  all  the  angles  and  the  side 
Er,  whence  E;z  will  be  given  ;  next,  in  the  triangle  AE?i 
will  be  given  (besides  the  right  angle)  both  the  legs  E/z 
and  EA,  whence  the  angle  EwA  is  given  ;  then  it  will  be, 
as  the  radius  to  the  sine  of  DH/z  or  Er;z  (:  :  DH  :  D;z  :  : 
DA  :  D;z)  so  is  the  sine  of  D72A  to  the  sine  of  DA?? ; 
whence  AD;z,  the  supplement  of  ACB,  is  also  given,  from 
which  all  the  rest  of  the  angles  in  the  figure  are  given  by 
addition  and  fiubtractioa  onl/. 


276 


The  Construction  of 


This  method  of  solving  the  problem,  it  may  be  observ- 
ed, requires  three  operations  by  the  sines  and  tangents ; 
but  the  same  thing  may  be  performed  by  two  proportions 
only ;  for  as  Er  :  AE  :  :  the  secant  of  rEn  :  the  tangent 
of  En  A  J  whence  all  the  rest  will  be  found  as  above. 

9 

PROBLEM  LXII. 

•  To  reduce  a  given  triangle  into  the  form  of  another^  or  to 
make  a  triangle  xvhich  shall  be  similar  to  one  triangle^  and 
equal  to  another. 

CONSTRUCTION. 

Upon  the  base  AB  of  the  triangle  ABC,  to  which  you 
would  make  another  triangle  equal,  describe  ADB  similar 

to    the    trian- 
D  gle  required; 

draw  CF  pa- 
rallel to  AB, 
meeting  AD 
inF;takeAE 
a  mean  pro- 
portional be- 
tween AD  and 
AF ;  and,  pa- 
rallel to  DB, 


G-  B 


draw  EG  ;  then  will  AGE  be  the  triangle  that  was  to  be 
constructed. 

DEMONSTRATION. 

Let  FR  and  DQ  be  perpendicular  to  AB ;  then  the 
triang.  ADB  :  triang.  ACB  :  :  DQ  :  FR  (schoL  Euc. 
1.  6.)  :  :  AD  :  AF  {Euc.  4.  6.)  :  :  AD^  :  AD  X  AF 
(JEwc.  1.  6.)  :  :  AD^  :  AE^  (by  construction)  :  :  triang. 
ADB  :  triang.  AEG  {Euc.  19.  6.).  Therefore  the  ante- 
cedents of  the  first  and  last  of  these  equal  ratios  being  the 
same,  the  consequents  ACB  and  AEG  must  necessarily 
be  equal.     ^.  E.  D. 

Method  of  Calculation. 

In  the  triangle  ADB  are  given  all  the  angles  and  the 
side  AB,  whence  AD  will  be  given  ;  next,  in  the  tri- 
angle AFR  will  be  given  all  the   angles  and  the  side 


Geometrical  Problems, 


377 


FR  (=CH),  whence  AF  will  be  given ;  and  then  AD 
and  AF  being  given,  AE  =  V  AD  x  AF  will  also  be 
given, 

PROBLEM  LXIIL 

To  find  a  point  in  a  given  triangle  ABC ^  from  xvhence 
right  lines  drawn  to  the  three  angular  points^  shall  divide 
the  whole  triangle  into  parts  (COA,  AOB,  BOC)  having 
the  same  ratio  one  to  another^  as  three  given  right  lines  m^ 
n,  and p^  respectively > 

CONSTRUCTION. 

In  CA  and  AB  produced,  if  need  be,  take  CE  and 
AF,  each  equal  to  m  +  n  +/?,  joining  E,  B,  and  F,  C  ; 
take  C^  =  w, 
Ac  =  n,  and 
draw  eh  and  cf 
parallel  to  EB 
and  CF,  meet- 
ing the  sides  of 
the  given  tri- 
angle in  h  and 
f\  draw  also 
hOi  and  /?  pa- 
rallel     to     AC 

and    AB,     and       ^      ^       -n  T"/^  a 

at    O,    the    in-       /^     ^       ^    ^   ;  -^ 

tersection  of  these  lines,  will  be  the  point  required. 


E 


DEMONSTRATION. 

Let  ^H  and  BD  be  perpendicular  to  AC.  The  trian- 
gles CBE,  Che^  as  also  CBD,  C^H,  are  similar;  there- 
fore, m  (C^)  :  m  +  72  +/?  (CE)  :  :  C^  :  CB  :  :  <5»H  :  BD 
:  :  the  triangle  AOC  :  triangle  ABC.  In  the  very  same 
manner  it  may  be  proved,  that  the  part  AOB  is  to  the 
whole  triangle  ABC,  as  n  tom-f-n-f/^>  whence  it  fol- 
lows, that  the  remaining  part  BOC  must  be  to  the  whole 
triangle,  as  /?  to  m  -f-  72  -|- /? ;  therefore  these  parts  are  to 
one  another  in  the  given  ratio  of  ?n,  n^  and  /?.     ^.  £.  D. 

3C 


378  The  Construction  of 

Method  of  Calculation. 

^"^^^     lm  +  7z+/^:n  ::  AC:  A/(QO), 
both  AQ  and  QO  will  be  given  from  thence  ;    then,  in 
the  triangle  AOQ,  will  be  given  two  sides,  and  the  in- 
cluded   angle,    from    which    every    thing    else    will    be 
known. 

PROBLEM  LXIV^ 

To  divide  a  given  trapezium  ABCD,  -whose  opposite  sides 
AB,  CD  are  parallel^  according  to  a  given  ratio^  by  a 
right  line  QN,  passing  through  a  given  point  P,  and  fall- 
ing  up07i  the  two  parallel  sides. 

CONSTRUCTION. 

Bisect  AD  in  G, 
■j^T  W  C X  and  draw  GH  pa- 
rallel to  AB  (or 
DC),  meeting  BC 
in  H  :  then  divide 
GH  in  M,  accord- 
ing to  the  given 
ratio,  and  through 
M  draw  PQN,  and 
the  thing  is  done. 

DEMONSTRATION. 
Draw  EMF  and  IHK  parallel  to  AD,  meeting  DC  and 
AB  in  E,  I,  K,  and  F. 

Because  of  the  parallel  lines,  we  have  GD  =  ME 
=  HI,  and  AG  =  FM  =  KH  ;  whence,  as  GD  is  = 
AG  (by  construction),  ME  will  be  =  FM,  and  HI  =  HK ; 
and  the  triangle  EMN  will  be  =  FMQ,  and  IHC  = 
BHK  (^Euc.  4.  1.)  ;  whence  it  appears  that  the  trapezium 
AQND  is  also  equal  to  the  parallelogram  DF,  and  the  tra- 
pezium QBCN  equal  to  the  parallelogram  FI ;  but  these 
parallelograms  are  to  one  another  as  their  bases,  or  as  GM 
to  MH  (^Euc.  1.6.);  therefore  GM  :  MH  :  :  ANQD  : 
QBCN.     ^E.D. 

Method  of  Calcidatioiu 
Whereas  AB  and   DC  are  parallel,  GH   is  an  arith- 
metical mean  between  them,  and  therefore  equal  to  half 


Geometrical  Problems*  o79 

their  sum»  Therefore,  as  the  whole  line  GH  and  the  ra- 
tio of  its  parts  GM,  MH  are  given,  the  parts  themselves 
will  also  be  given. 

PROBLEM  LXV. 

To  cut  off  from  a  given  trapezium  ABCD,  -whose  vp- 
posite  sides  AB^  CD  are  parallel^  a  part  AQND  equal  to 
a  rectangle  given^  by  a  right  line  passing  through  a  given 
point  P,  andfallifig  upon  the  two  parallel  sides,  ij^ee  the 
figure  to  the  last  problem. 

CONSTRUCTION. 

Bisect  AD  in  G,  and  draw  GH  parallel  to  AB  ;  vipon 
AD  {by  Euc.  45.  1.)  describe  the  parallelogram  ADEF 
equal  to  the  rectangle  given,  and  through  the  intersec- 
tion of  GH  and  EF  draw  PQN,  and  the  thing  is  done. 
The  demonstration  whereof  is  manifest  from  the  preceding 
problem. 

PROBLEM  LXVL 

To  divide  a  given  trapeziu7n  ABCD,  whose  sides  AB 
and  DC  are  parallel^  into  tzvo  equal  parts^  by  a  rights  line 
parallel  to  those  sides. 

CONSTRUCTION. 

Produce      AD  ^. 

and  BC    till  they  7  \ 

meet    in    H,    and  /       \ 

make    AG    equal  /  \   . 

and  perpendicular  /  ^\ 

toHD;drawHG,  T^/        E/^     \ 

andbisect  the  same  /  "^'X/N^ 

with  the  perpendi-  /p.  ^x^'n      \ 

cular  PQ  =  HP ;  //\  "^"-AX  \ 

join  H,  Q,  and  in  //   J\ ni\'         \ 

HA     take      HE,        A^---x ^^LX A 

equal  to  HQ,  and        "^  *-  ,       ^  "^ 

parallel  to  AB  draw  EF,  and  the  thing  is  done. 
DEMONSTRATION. 

Since  HE^    (=  HQ^  ^  hP^  +  PQ^   =   2HP2  t=r: 
HG^        HA2  +  AG^        HA2  +  HD^    . 
-^  = =  1 )    IS  an  anth- 


380 


The  Construetion  of 


metical  mean  between  HA^  and  HD%  it  is  evident  that 
the  triangle  HEF  will  also  be  an  arithmetical  mean  be- 
tween the  triangles  H  AB  and  HDC  (or  ABFE  =  EFCD) ; 
because  those  triangles,  being  similar,  are  to  one  another 
as  (HE%  HA^,  HD^)  the  squares  of  their  homologous 
sides.     4L'  -^-  ^* 

Method  of  Calculation. 
Since  all  the  sides  and  angles  of  the  trapezium  are  sup- 
posed given,  the  side  CD  and  all  the  angles  of  the  trian- 
gle HDC  will  be  given;    therefore  HD  and  AH  will  be 

given.  But  the  same  thing  may  be  had  without  the  an- 
gles ;  for,  since  DC  is  parallel  to  AB,  we  have  AB  —  DC 
:  AD  :  :  DC  :  HD  ;  whence  HE  will  be  given,  as  be- 
fore. 

PROBLEM  LXVII. 


known :    whence  HE, 


firiD^  4-  HA2 


will  also  be 


To  divide  a  given  trapezium  ABCD  according  to  a 
given  ratio ^  by  a  right  line  LH  cutting  the  opposite  sides 
AC,  BD  in  given  angles, 

CONSTRUCTION. 

Produce  the  said  opposite  sides  till  they  meet  in  E ; 


draw  AD,    and 


BX 


CF  parallel  to  it,  meeting  BE   in   F ; 

divide  BF  in 
G,  accord- 
ing to  the 
given  ratio ; 
and,  having 
made  EAK 
equal  to  the 
given  angle 
which  LH 
is     to    make 


wi-h  AC,  take  EH  a  mean  proportional  between  EG  and 

EK  ;  tiien  draw  HL  parallel  to  AK,  and  the  thing  is  done. 

DEMONSTRATION. 

By  construction,  EG  :   EH  :  :  EH  :  EK  :   :  EL  : 

EA  {Eiic.  5.  6.)  ,•    whence  it  follows  that  EG  X  EA  = 


Geometrical  Problems. 


381 


EH  X  EL^  and  consequently  that  the  triangles  EHL 
and  EAG  are  also  equal  to  each  other  (£i(c.  15.  6.), 
from  which  taking  away  EDC,  common,  the  remainders 
CDHL  and  CDGA  will  be  likewise  equal,  and  conse- 
quently ALHB  =  AGB,  being  the  dift'erences  between 
those  remainders  and  ACDB.  But  the  triangle  ADF  is 
=  ACD,  standing  upon  the  same  base  AD  and  between 
the  same  parallels  ;  therefore  (by  adding  AGD,  common) 
AGF  is  also  =  CDGA  (=  CDHL)  ;  but  AGF  (CDHL) 
:  AGB  (ALHB)  :  :  GF  :  GB  {Euc.  1.  6.).     ^  E.  D. 

Method  of  Calculation. 
In  the  triangles  ABE  and  ABK  are  given  all  the 
angles  and  the  side  AB,  whence  BE,  BK,  and  EC  will 
be  known ;  then  in  the  triangle  EFC  will  be  given 
all  the  angles  and  the  side  CE,  whence  EF,  and  from 
thence  FG,  and  EG,  will  be  known;  lastly,  from  the 
known  values  of  EK,  EG,  and  EF,  the  value  of  FH  (  = 
VEG  X  E"K  —  EF)  will  be  found. 


PROBLEM  LXVIIL 

Two  right  li?ies  AG  a7id  AH,  7neeti7ig  in  a  point  A, 
keing  given  by  position  ;  it  is  required  to  draw  a  right  line 
TzP  to  cut  those  lints  in  given  angles^  so  that  the  triangle 
KnV^  formed  from  thence^  mau  be  equal  to  a  given  square 
ABCD. 

CONSTRUCTION. 

Let  the  angle  ABE  be  equal  to  the  given  angle  APn, 
and  let  BE 
meet  AG  in 
E;  drawEF 
perpendicu- 
lar to  AH, 
make  BQ 
equal  2EF, 
and  upon 
AQdescribe 

the  semicircle  AmQ,  cutting  BC  in  rn  ;  draw  mn  parallel 
to  AH,  meeting  AG  in  w,  and  n?  parallel  to  EB,  and 
A^P  will  be  the  triangle  required. 


382 


The  Construction  of 


DEMONSTRATION. 

The  triangles  AEB  arid  A;zP,  being  similar,  are  to  one 
another  as  the  squares  of  their  perpendicular  heights  EF 
and  mB  (/zS) :  but  mW  is  =  BQ  x  AB  =  2EF  x  AB 
therefore  it  will  be,  as  the  triangle  AEB  (EF  x  J-AB) 
the  triangle  A;zP  :  :  EF^  :  2EF  x  AB  :  :  EF  :  2AB  : 
EF  X  |AB  :  AB2  [Euc.  1.  6.);  wherefore,  the  antece- 
dents being  the  same,  the  consequents  must  necessarily  be 
equal,  that  is,  An?  =  ABCD.  ^.  E.  D. 
Method  of  Calculation. 

In  the  triangle  ABE  are  given  all  the  angles  and  the 
side  AB,  whence  EF  will  be  given,  and  consequently 
S;z  (=  VAB  X  2EF)  ;  whence  AP  and  An  are  also 
given. 

LEMMA. 

If  from  any  point  C,  in  one  side  of  a  plane  angle  KAL, 
a  right  line  CB  be  draxvn^  cutting  both  sides  AK,  AL  in 
equal  angles  (ACB,  ABC);  and  from  any  other  point  D 
in  the  same  side  AK  another  right  line  be  drawn^  to  cut  off 
an  area  ADE  equal  to  the  area  ABC ;  Isay^  that  DE  will 
be  greater  than  CB. 

DEMONSTRATION. 
Complete  the  parallelogram  DCBG,  and  join  B,  D,  and 

in  BG  (produced  if  need  be)  take  BF  =  BE,  and  draw 

FD. 

Since  the  triangles  ABC,  AED  are  equal,  by  suppo- 
sition, and  have  one 
angle.  A,  common, 
therefore  will  AD  : 
AC  :  :  AB  (AC)  : 
AE  {Euc.  15.  6.), 
and  consequently 
AD  -f  AE  greater 
than  AC  -f.  AB 
( Euc.  25.  5.  )  ; 
whence  it  is  mani- 
fest that  CD  must 
be  greater  than  EB,: 

or  BG  than  BF.      Moreover^  because  the  angle  ABC 


Geometrical  Problems. 


383 


(  =  ACB  =  CDG)  is  =:  CxBC,  it  will  be  greater  than 
GBD,  which  is  but  a  part  of  GBC  ;  and  therefore  ABD 
must,  evidently,  be  greater  than  GBD  ;  wherefore,  seeing 
BF  and  BE  are  equal,  and  that  DB  is  common  to  both 
the  triangles  DBE,  DBF,  it  is  manifest  that  DE  is 
greater  than  DF  {Euc.  19.  1.)  ;  but  DF  is  greater  than 
jyO  (by  the  same)^  because  the  angle  DGF  (DCBj,  be- 
ing obtuse,  is  greater  than  GFD,  which  must  be  acute 
{Euc,  32.  1.) :  consequently  DE  is  greater  than  DG,  or 
its  equ^l  CB.     ^  E.  B. 


PROBLEM  LXIX. 

From  a  given  polygon  ABCDEF,  to  cut  off* a  given  area 
AFEIK,  by  the  shortest  right  line^  Kl^  possible. 

CONSTRUCTION. 

Let  the  given  area  to  be  cut  off  be  represented  by 
the  rectangle  LMNO  ;  and  let  the  sides  AB  and  DE, 
by  which  the  dividing  line  is  terminated,  be  produced 
till  they  meet  in  G ;  make  upon  OL  ^by  Euc.  45.  1 .)  a 
rectangle  OQ  equal  to  AFEG,  and  let  a  square  GSTV 


be  constituted  (by  Euc.  14.  2.)  equal  to  the  whole 
rectangle  QN :  bisect  the  angle  BGD  by  the  right  line 
GH,  and  make  GR  perpendicular  to  GH ;     ajid  draw 


384 


The  Construction  of 


KI,  by  the  last  proMem,  parallel  to  RG,  so  as  to  form  the 
triangle  KGI  equal  to  the  square  GSTV,  and  the  thing 
is  done. 

DEMONSTRATION. 

Since,  by  construction,  KGI  (  =  GSTV)  =  QN,  let 
AFEG  =  OQ  be  taken  away,  and  there  will  remain 
AFEIK  =  LN.  Moreover,  since  the  angle  HGI  is  :^ 
HGK,  and  the  angle  IHG  (HGR)  a  right  one,  the  angles 
I  and  K  are  equal ;  and  therefore,  by  the  preceding  lemma, 
IK  is  the  shortest  right  line  that  can  possibly  be  drawn  to 
cut  off  the  same  area.     ^.  -E,  D. 

Method  of  Calculation* 

Let  the  area  of  the  figure  AFEG  be  found,  by  divid- 
ing it  into  triangles  AFG,  EFG,  and  let  this  area  be  add- 
ed to  the  given  area  to  be  cut  off,  and  then,  the  square 
root  of  the  sum  being  extracted,  you  will  have  GS  the  side 
of  the  square  .GT  ;  from  whence  GI  will  be  determined, 
as  in  the  last  problem. 

Note.  In  the  same  manner  may  a  given  area  be  cut  off 
by  a  right  line  making  any  given  angles  with  the  opposite 
sides. 

PROBLEM  LXX. 

Through  a  given  point  P,  to  draw  a  right  line  PED  to 
cut  txvo  right  lines  AB,  AC  given  in  position^  so  that  the 
triangle  AJ^^  ^formed from  thence^  may  be  of  a  given  mag^ 
nitude* 

CONSTRUCTION. 

Draw   PFH  parallel  to  AB,  intersecting  AC  in  F; 


DB 


Geometrical  Problems.  385 

and  upon  AF  let  a  parallelogram  AFHI  be  constituted 
equal  to  the  given  area  of  the  triangle  ;  make  IK  perpen- 
dicular to  A  I,  and  equal  to  FP  ;  and,  from  the  point  K, 
to  AB,  apply  KD  =  PH  ;  then  draw  DPE,  and  the  thing 
is  done. 

DEMONSTRATION. 
Supposing  M  to  be  the  intersection  of  DF  and  IH,  it 
is  evident,  because  of  the  parallel  lines,  that  all  the  three 
triangles  PHM,  PFE,  and  MDI  are  equiangular  ;  there- 
fore, all  equiangular  triangles  being  in  proportion  as  the 
squares  of  their  homologous  sides,  and  the  sum  of  the 
squares  of  PF  (IK)  and  DI  being  equal  to  the  square  of 
PH  (KD),  by  construction  and  Euc.  47.  1.  it  is  evident 
that  the  sum  of  the  triangles  PFE  and  DMI  is  =  the 
triangle  PHM;  to  which  equal  quantities,  vi Jig*  1,  let 
AFPMI  be  added,  so  shall  ADE  be  likewise  equal  to 
AFHI :  but,  infg.  2,  let  PFE  be  taken  from  PHM,  and 
there  will  remain  EFHM  =  DMI;  to  which  adding 
AIME,  .we  have  AFHI  =  ADE,  as  before.     ^  E.  D. 

Method  of  Calculation. 
By  dividing  the  given  area  by  the  given  height  of  the 
point  P  above  AB,  the  base  AI  of  the  parallelogram  AFHI 
will  be  known,  and  consequently  PH  (  =  KD)  ;  whence 
DI  (=  VKD^  —  PF^)  will  likewise  become  known. 
This  problem,  it  may  be  observed,  becomes  impossible 
when  KD  (PH)  is  less  than  KI  (PF)  ;  which  can  only 
happen,  in  case  1,  when  the  given  area  is  less  than  a  pa- 
rallelogram under  AF  and  FP. 

PROBLEM  LXXI. 

To  cut  off  from  a  given  polygon  BCIFGH,  a  part 
EDBHG  equal  to  a  given  rectangle  KL,  by  a  right  line 
^D  passing  through  a  given  point  P. 

CONSTRUCTION. 

Let  the  sides  of  the  polygon  CB  and  FG,  which  the 
dividing  Ime  ED  falls  upon,  be  produced,  till  they  meet 
in  A  ;    upon  ML  (by  Euc.  45.  1.)  make  the  rectangle 

3D 


386 


The  Construction  of 


MN  equal  to  AGHB,  and,  by  the  last  problem,  let  ED 
be  so  drawn  through  the  given  point  P,  that  the  triangle 
A  ED,  formed  from  thence,  may  be  equal  to  the  whole 


li  s 

rectangle  KN  ;  then  will  EDBHG  be  equal  to  KL : 
for  since  AED  is  =  KN,  let  the  equal  quantities  AGHB 
and  MN  be  taken  away,  and  there  will  remain  EDBHG 
=  KL. 

Method  of  Calculation* 
Let  the  area  of  the  figure  AGHB  be  found,  by  divid- 
mg  it  into  triangles,  and  l^t  this  area  be  added  to  the  area 
given,  and  the  sum  will  be  equal  to  the  area  AED,  or  the 
rectangle  KN  ;  from  whence  AD  will  be  found,  as  in  the 
last  problem. 

PROBLEM  LXXIL 

HaviJig  the  base^  the  vertical  angle^  and  the  length  of  the 
line  bisecting  that  angle  and  terminating  in  the  base^  to 
describe  the  triangle. 

i  CONSTRUCTION. 

Upon  the  given  base  AB  let  a  segment  of  a  circle 
ACB  be  described  [by  problem  4)  to  contain  the  given 
angle,  and,  having  completed  the  whole  circle,  from 
O,  the  center  thereof,  perpendicular  to  AB,  let  the  ra- 
dius OE  be  drawn  j    draw  EB,  and  make  BG  perpeit- 


Geometrical  Problems, 


385 


dicular  thereto,    and  equal  to  half   the  given  bisecting 

line,  and  from  G,  as  ^ -.s^  r; 

a  center,  with  the  ra- 
dius GB,  let  a  circle 
BHF  be  described, 
intersecting  EG 

(when  drawn)  in  F 
and  H  ;  from  E  to 
ABdrawED  =  EF, 
and  let  the  same  be 
produced  to  meet  the 

circumference  in  C  ;  join  A,  C,  and  B,  C  ;  so  shall  ABC 
be  the  triangle  required. 

DEMONSTRATION. 

The  triangles  CBE  and  BDE  are  similar,  because 
the  angle  BEC  is  common  to  both,  and  the  angles  BCE 
and  DBE  stand  upon  equal  arches  BE  and  AE  ;  there- 
fore EC  :  EB  :  :  EB  :  ED,  and  consequendy  ED  x  EC 
=  EB2:  but  {by  Euc.  36.  3.)  EB^  =  EF  X  EH  =  ED 
X  EH  (b7j  comtruction).  Hence  EDxEC  =  EDx 
EH,  and  consequently  EC  =  EH  ;  from  which  taking 
away  the  equal  quantities  ED  and  EF,  there  remains 
DC  =  FH  =  the  given  line  bisecting  the  vertical  angle 
(^by  construction)  :  and  it  is  evident  that  DC  bisects  the 
angle  ACB,  since  ACD  and  BCD  stand  upon  equal  arches 
AE  and  EB.     ^  E.  D. 

Method  of  Calculation. 
If  BE  be  considered  as  a  radius,  BR  Q  AB)  will  be  the 
co-sine  of  the  angle  EBR,  and  BG  the  tangent  of  BEG  ; 
therefore  BR  :  BG  (or  AB  :  DC)  : :  co-sine  EBR  (ACE) 
:  tang.  BEG,  whose  half  complement  EHB  is  likewise  giv- 
en from  hence  :  then  the  angle  HB^  (supposing  EB  pro- 
duced to  b^  being  the  complement  of  EHB,  we  shall  have 
tang.  EHB  :  rad.  (  :  :  sine  EHB  :  co-sine  EHB  :  :  BE  : 
EH  :  :  EB  :  EC)  :  :  sine  ECB  :  sine  CBE  =  sine  EDB 
=  co-sine  OED,  half  the  difference  of  the  angles  (ABC 
and  B  AC)  at  the  base. 


388 


The  Construction  of 


PROBLEM  LXXIIL 

Having  given  the  two  opposite  sides  ab^  cd^  the  two  diago- 
nals aCj  bd^  and  also  the  angle  aeb  in  which  they  intersect 
each  other  ;  to  describe  the  trapezium* 

CONSTRUCTION. 

In  the  indefinite  line  BP  take  BD  equal  to  bd^  and  make 
the  angle  DBF  equal  to  the  given  angle  aeb^  and  BF  =  ere; 
also  from  the  centers  D  and  F,  with  the  radii  dc  and  ab^ 


let  two  arches  mQ.n  and  r^s  be  described  intersecting 
each  other  in  C  ;  join  D,  C,  and  F,  C,  and  make  BA 
equal  and  parallel  to  FC  ;  then  draV  AD,  AC,  and  BC, 
and  the  thing  is  done. 

DEMONSTRATION. 

Since  {by  construction^  AB  is  equal  and  parallel  to  CF, 
therefore  will  AC  be  equal  and  parallel  to  BF  {Euc»  33. 
1.),  and  consequently  the  angle  AEB  {Euc.  29.  1.)  = 
DBF  =  aeb.     ^.  E.  D, 

Method  of  Calculation, 

Join  D,  F  ;  then  in  the  triangle  DBF  will  be  given  two 
sides  DB,  BF,  and  the  angle  included  -,  whence  the  angle 
BFD  and  the  side  DF  will  be  known  ;  then  in  the  trian- 
gle DFC  will  be  given  all  the  three  sides ;  whence  the 
angle  DFC  will  be  known,  from  which  BFC  (BFD  — 
DFC)  =  BAC  will  also  be  known. 


Geometrical  Problems, 


'  389 


PROBLEM  LXXIV. 

Having  given  the  two  diagonals  and  all  the  angles^  to 
describe  the  trapezium, 

CONSTRUCTION. 

Assume  AS  at  pleasure ;  and,  having  produced  the 
same  both  ways,  make  the  angles  QAC,  RBC  equal, 
respectively,  to  two  opposite  angles  a  and  e  of  the  tra- 
pezium;  moreover,  make  ACF  equal  to  ace^  one  of 
the  remaining  angles  ;  and  from  F,  the  intersection  of 
AF  and  BQ,  take   FG  =  the  given  diagonal  dc^    and 


q:f 


J)  OP 


draw  GH  parallel  to  CB,  meeting  FB  in  H.  Then  from 
A  and  B  {by  the  lemma^  p.  336)  let  two  lines  AE  and  BE 
be  drawn  to  meet  in  FC,  so  as  to  be  in  the  given  ratio  of 
tz^  to  FH  ;  in  AE  take  AN  =  ae^  and/  draw  NM  paral- 
lel to  FC,  meeting  AC  in  M  ;  lastly,  draw  NP  making 
the  angle  MNP  ==  ced^  and  meeting  FB  in  P  ;  so  shall 
A  MNP  be  the  true  figure  required. 

DEMONSTRATION. 

Let  ED  be  parallel  to  NP,  and  let  DC  and  PM  be 
drawn. 

It  is  evident,  by  construction,  that  the  diagonal  AN, 
and  all  the  angles  of  the  trapezium,  are  equal  to  the 
respective  given  ones ;  it  therefore  remains  only  to 
prove  that  PM  is  equal  to  the  other  given  diagonal  dc. 
Now  the  angle  RBC  being  =  CED  {by  coiist ruction)^ 
the  circumference  of  a  circle  may  be  described  through 
all  the  four  angular  points  of  the  trapezium  BCED  ; 
and  so  the  triangles  FBE  and  FCD  (as  both  the  angles 
FBE  and  FCD  stand  upon  the  same  chord  ED)  will 


390  The  Construction  of 

be  similar ;  and  consequently  BE  :  DC  ( :  :  FB  :  FC) 
:  :  FH  :  FG  {dc).  But  (by  construction)  AE  :  BE  :  : 
ae  :  FH ;  therefore,  by  compounding  these  two  pro- 
portions, we  have  AE  :  DC  :  :  ae  :  dc ;  but  (because  of 
the  similar  figures  ADEC,  APNM)  we  also  have  AE  : 
DC  :  :  AN  (ae)  :  PM  ;  and  consequently  PM  =  dc. 
^.  E.  D. 

Method  of  Calculation. 

All  the  angles  of  the  triangles  ABC,  FAC,  and  FBC 
being  given,  we  shall  have  sine  ACB  x  sine  F  :  sine 
ABC  X  sine  ACF  :  :  AB  :  AF  ;  and  sine  FHG  (FBC) 
:  sine  FGH  (FCB)  :  :  FG  (^c)  :  FH ;  whence  AF  and 
FH  are  known. 

Find  AK  =  ^^^  and  KO  =  ^I^ :  which 
FH  +ae  h  ii  —  ae 

last  is  equal  to  (OE)  the  radius  of  the  circle  determining 
the  point  E  {see  the  aforesaid  lemma).  Therefore,  in  the 
triangle  F'OE  are  given  two  sides  FO  and  OE,  besides 
the  angle  F  ;  whence  the  angle  FOE  will  be  given  ;  then 
in  the  triangle  AOE  will  be  given  OA,  OE,  and  the  in- 
cluded angle  ;  whence  the  angle  OAE,  which  the  diago- 
nal AN  makes  with  the  side  AP,  will  be  known,  and  from 
thence  every  thing  else  required. 

This  problem,  as  the  circle  described  frOm  O  cuts  FC 
in  two  points,  admits  of  two  different  solutions  (except, 
only,  when  FC  touches  the  circle).  If  the  circle  neither 
puts  nor  touches  that  line,  the  problem  will  be  impossible  ; 
the  limits  of  the  ratio  of  AE  to  BE  (and  consequently  of 
ae  to  dc)  growing  narrower  and  narrower,  as  AB  be- 
comes less  and  less,  with  respect  to  AC,  or  according 
as  the  sum  of  the  opposite  angles  (a  -f  e  =  QAC  -f-  RBC) 
approaches  nearer  and  nearer  to  two  right  angles  ;  so  that, 
at  last  (supposing  AC  and  BC  to  coincide),  AE  and  BE 
will  be,  every  where,  in  the  ratio  of  equality  ;  therefore 
cd  can  here  have  only  one  particular  ratio  to  ae  ;  and  the 
diagonal  ANE  may  be  drawn  at  pleasure,  the  problem 
being,  in  this  case,  indeterminate. 


Geometrical  Problems. 


391 


PROBLEM  LXXV. 

Supposing  the  right  lines  m^  n^p  to  represent  the  lengths 
of  three  staves  erected  perpendicular  to  the  horizon^  in  the 
given  points  A,  B,  C  ;  to  find  a  point  P,  in  the  plane  of 
the  horizon  ABC,  equally  remote  from  the  top  of  each 
staff. 

CONSTRUCTION. 

Join  A,  B,  and  B,  C,  and  make  AE  and  BF  perpen- 
dicular to  AB  ;  also  make  BG  and  CH  perpendicular 
to  BC,  and  let  AE  be  taken  =  m,  CH  =/?,  and  BF  and 
BG  each  =  w;    draw  EF  and  GH,  which  bisect  by  the 


-£  \ 


r 

'ft 

G 

£.— ■ 

•-•••■"  I 

perpendiculars  LN  and  IK,  cutting  AB  and  CB  in  N 
and  K  ;  make  KP  and  NP  perpendicular  to  BC  and  BA, 
and  the  intersection  Y  of  those  perpendiculars  will  be  the 
point  required. 

DEMONSTRATION. 
Conceive  the  planes  AEFB  arid  BCHG  to  be  turned 
up,  so  as  to  stand  lerpendicular  to  the  plane  of  the  ho- 
rizon ABC  and  ircerser.t  it  in  the  right  lines  AB  and 
BC  ;  then,  becaup  BF  and  BG  arc  equal  to  each  other, 
and   perpendicul}^    to    the  plane   of  the  horizon,  it  is 


392  The  Construction  of 

evident  that  the  points  F  and  G  must  coincide,  and  that 
AE,  BG  (BF),  and  CH  will,  represent  the  true  position 
of  the  staves  :  suppose  KG,  KH,  PG,  PH,  PE,  and 
PF  to  be  now  drawn ;  then,  since  (by  construction^  GI 
=  HI,  and  the  angle  GIK  =  HIK  ;  therefore  is  GK 
=  HK  [Euc.  4.  1.):  moreover,  since  KP  is  (by  con- 
struction) perpendicular  to  BC,  it  will  also  be  perpendi- 
cular to  the  plane  BCHG,  and  consequently  the  angles 
PKG  and  PKH  both  right  angles:  therefore,  seeing 
the  two  triangles  GKP,  HKP  have  two  sides  and  an 
included  angle  equal,  the  remaining  sides  PG  and  PH 
must  likewise  be  equal  (^Euc.  4.  1.).  After  the  very 
same  manner  it  is  proved  that  PF  (or  PG)  is  equal  t© 
EP.     ^  E.  D. 

Method  bf  Calculation* 
Draw  Ir  perpendicular,  and  H^  parallel  to  BC  ;  then, 
by  reason  of  the  similar  triangles  H^G  and  IrK,  it  will 

be  as  BC  (%)  :  BG  —  CH  (G^)  :  :  5£±£5  (I^) 


^  BG  —  CH  X  BG  +  CH        ,  .  ,        ,  , 

:  Kr  =    =rp; ;    which  subtracted 

2BC 

from  Br  (  =  |BC)  gives  BK  :  and  in  the  same  man- 
ner will  BN  be  found ;  then  in  the  trapezium  KBNP 
will  be  given  all  the  angles  and  the  two  sides  BK  and 
BN ;  from  whence  the  remaining  sides,  &c.  may  be 
easily  determined. 

PROBLEM  LXXVI. 

The  base^  the  perpendicular^  and  the  difference  of  the  sides 
being  groen^  to  determine  the  triangle. 

CONSTRUCTION. 

Bisect  the  base  AB  in  C,  and  in  it  take  CD  a  third  pro- 
portional to  2AB  ^nd  the  given  difference  of  the  sides 
MN ;  erect  DE  equal  to  the  giren  perpendicular,  and 
draw  EK  parallel  to  AB,  and  take  therein  EF  =  MN ; 
draw  E  AG,  to  which,  from  F,  apply  FG  =  AB  ;  draw 
AH  parallel  to  FG,  meeting  EK  In  \ :  then  draw  BH, 
and  the  thing  is  done. 


Geometrical  Problems, 


39G 


DEMONSTRATION. 

By  reason  of  the  parallel  lines,  FG  (AB)  :  FE  (MN) 
: :  AH  :  EH  (DP)  ;  therefore  AB  x  DP  =  AH  x  MN, 
or  2AB  X  DP  =  2AH 

X  MN;  to  which  last 
equal  quantities  adding 
2ABxCD  =  MN%(^y 
construction)  we  have 
2AB  X  CP  =  2AH  X 
MN  X  ^•-  MN2  ;  but 
2AB  X  CP  is  =  BH2 
—  AH^  (by  a  known 
property  of  triangles)  ; 
therefore   BH^  _  AH^ 

=  2AHxMN'+MN%  

or  BH^  =  KW  +  2 AH  x  MN  +  MN^  =  AH  +  U^Y 
{Euc.  4.  2.),  consequently  BH  =  AH  +  MN.     ^  E.  D. 

Method  of  Calculation. 
In  the  right-angled  triangle  ADE   we  have  DE  and 
/  MN2\ 

AD  f  =  |AB  —  j^)  >  whence  the  angle  DAE  (FEG) 

will  be  found  ;  then  in  the  triangle  EFG  will  be  given  two 
sides  and  one  angle,  from  which  the  angle  GFK  (  =  BAH) 
will  also  be  known. 


PROBLEM  LXXVIL 

The  base^  the  perpendicular^  and  the  sum  of  the  two  sides 
being'  given^  to  describe  the  triangle* 

CONSTRUCTION. 

Bisect  the  base  AB  in  C,  and  in  it  produced  take  CD 
a  third  proportional  to  2AB,  and  the  sum  of  the  sides, 
MN  ;  erect  DE  equal  to  the  given  perpendicular,  and 
draw  HE  parallel  to  AB,  and  take  therein  EF  =  MN; 
draw  EAG,  to  which,  from  F,  apply  FG  =  AB  ;  draw 
AH  parallel  to  FG,  meeting  EF  in  H ;  then  draw  BH. 
and  the  thing  is  done. 
/  3E 


m^ 


The  Gonstructton  of 


DEMONSTRATION. 

Because  of  the  parallel  lines,  FG  (AB)  :  FE  (MN) 
2  :  AH  :  EH  (DP)  ;  and  therefore  2lMN  x  AH  = 
2AB  X  DP;  which  equal  quantities  being  subtracted 
from  MN^  =  2AB  x  CD,  {by  construction)  there  wiQ 

j^r       " '  j\i 


remain  MN^  —  2MN  x  AH  =  2AB  x  CP  =  BH^  _ 
AH^  ;  whence,  by  adding  AH^  to  each,  we  have 
MN^  —  2MN  X  AH  +  AH2  =  BHS  that  is, 
MN_AHT=:  BH^  therefore  MN  —  AH  =  BH, 
or  MN  =  BH  +  AH.     ^.  E.  D. 

Method  of  Calculation. 
In  the  triangle  AED  are  given  (besides  the  right  an- 
gle) both  the  legs  ;  whence  the  angle  DAE  (=  FEG) 
will  be  given  ;  then  in  the  triangle  FEG  one  angle  and 
two  sides  will  be  known,  from  which  the  angle  EFG 
(  =  BAH)  will  be  determined. 


PROBLEM  LXXVIIL 

The  difference  of  the  two  sides ^  the  perpendicular^  and 
the  vertical  angle  being  groen^  to  determine  the  triangle. 

CONSTRUCTION. 

Upon  the  indefinite  line  FEQ  erect  the  given  perpen- 
dicular DC,  making  the  angle  DCE  =  half  the  given 
angle;  let  EF,  expressing  the  given  difference  of  the 
sides,  be  bisected  by  the  perpendicular  GI,  meeting 
EC  in  I ;  also  let  EC  be  bisected  in  H,  and  make  EK 
perpendicular  to  CE,    and    equal  to  EI;     and,   having 


Geometrical  Problems.  „„„ 

drawn  HK    taVe  ht     •     rrr, 

from  L  to  FQ  appj^'il^B  -^^ '"!f  ^^  -^^"^  thereto ; 
i^i^7-'^«-EK,  and  join  C,B;  also 


draw  EM,  maHnrv  +1,1^  "^o  DEC,  and 

cuui.,  CB  rS?  tt^M^EF  apply  CA  = 
CM,  so  shall  ACB  be  the  tir^  >^q^^recl. 

Upon  EM  le^  ^-^  ^^^  perpendicular  CN,  and 
join  L,  M,  p.id  F.  J»  Now,  LB^  =  EK^  C^^y  --- 
^truction')  ^MTCm.  X  HK  —  f^  V^^^^-  ^-  ^0  == 
inriTcH  N>  HL  —  HK  0>«/  construction)  =  CL  X  EL; 
whencf-  rf^^  -"^  •  •  ^^  •  ^L  5  therefore  the  triangles 
^J^ii  and  ELB  must  be  equiangular  [Euc.  6.  6.),  and 
consequently  LBM  =  LEB  =  CED  =  CEM  {by  con- 
struction).  Therefore,  since  the  external  angle  CEM 
of  the  trapezium  LEMB,  is  equal  to  the  opposite  internal 
angle  B,  the  circumference  of  a  circle  will  pass  through 
all  the  four  angular  points  ;  and  consequently  the  an- 
gle LMB  will  be  =  LEB,  both  standing  upon  the 
same  chord  LB  ;  but  it  is  prov^ed  that  LBM  is  =  LEB  j 
therefore  LMB  =  LBM  =  FEI ;  and  so  the  triangles 
BLM  and  EIF,  being  isosceles,  and  having  LMB  =  EFI, 
and  also  LB  =  EI  (by  construction)^  they  will  be  equal 
in  all  respects,  and  consequently  BM  =  EF  ;  whence 
BC  —  AC  r  =  BC  —  CM  ==:  BM)  =  EF,  the  given 


396 


The  Construction  of 


difference  {by  construction^.  Moreover,  CEN  being  = 
CED  {by  construction^^  CN  will  be  =  CD;  and  so 
CM  being  =  CA,  ACD  will  be  =  MCN,  to  which  add- 
ing DCM,  common,  we  have  ACB  =  DCN  =  2DCE. 

Method  of  Calculation. 
Seeing  EG  and  EH  are  the  sine  and  tangent  of  EIG 
and  EKH,  to  the  equal  radii  EI  and  EK,  it  will  therefore 
be  EG  :  EH  (or  EF  :  EC)  :  :  sine  EIG  (ECD)  :  tan- 
gent EKH.  But  EC  :  CD  :  :  the  radius  :  co-sine  ECD  ; 
whence,  by  compounding  these  proportions,  EF  : 
CD  :  :  radius  x  sine   ECD  :  co-sine  ECD  x  tangent 

^__^^       radius  x  sine  ECD  ,       ^  ^  t^^tw     *. 

KH  :  :  — (  =  tangent  ECDj  :  tangent 

■^  co-sinc  jlCD 

will  u  fj,oj^  ^hich  EKL,  half  the  complement  of  EKH 
^'^^  ( :  .  (riven  ;  then  it  will  be,  as  the  radius  :  tangent 
:  sine  LBE  .^EL  :  :  LB  :  EL)  :  :  sine  LEB  (CED) 
words,  give  the  Fc&)  ;    which  proportions,  expressed   in 

As  the  difference  om%  theorem  : 
the  tangent  of  half  the  veH^^^  ^^  ^^  the  perpendicular^  so  is 
gle  ;  and  as  the  radius  is  to\ff'^S^^  ^^  ^^^^  tangent  of  an  an- 
ment  of  this  angle^  so  is  the  con^^S^^^^  ^f  ^^^^f  ^^^^  comple- 
to  the  sine  of  ha  f  the  diference  oft^^^^lf^^^^  vertical  angle 


r/z< 


^f^c  perpendicular,  the  differ 
^g^^ence  of  the  angles  :t  £^^^^ 


*  mi^U^  dt  the  base* 

"■^o^rxM  LXXIX. 


''nine  the  triangle. 


(angles  at  the  base  heint, 


g^n,  i^'iJt 


CJ)NSTRUCTI0N. 

l^tt  a  triangle  ABC 
be  constructed    by  the 
^ast problem.yvhose  per-- 
pendicular    and     diffe- 
rence of  the  sides  shall 
be  the  same  with  those 
given,  and  whereof  the 
vertical  angle  ACB  is 
also  equal  to  the  given 
difference    of    angles; 


Geometrical  Problems. 


397 


then  upon  C,  as  a  center,  with  the  radius  CB,  let  an  arch 
be  described,  intersecting  AB,  produced,  in  D  ;  join  C, 
D,  and  ACD  will  be  the  triangle  required.  For  CD  be- 
ing =  CB,  the  angle  CDB  will  also  be  =  CBD  =  A  + 
BCA  {Euc.  32.  1.}.  The  method  of  calculation  is  als© 
the  same  as  in  the  preceding  problem. 


PROBLEM  LXXX. 

The  perpendicular^  the.  sum  of  the  two  sides ^  and  the  ver- 
tical angle  being  given^  to  describe  the  triangle. 

CONSTRUCTION. 

Upon  AB,  the  given  sum  of  the  two  sides,  erect  AC 
equal  to  the  given  perpendicular ;  and  make  the  angle 
ACH  equal  to  the  complement  of  half  the  given  an- 
gle ;     upon  AB  (by  prob.   72)   let  a  triangle   ABF  be 


constituted,  whose  vertical  angle,  AFB,  shall  be  equal  to 
the  given  one,  and  whereof  the  bisecting  line  FE  (ter- 
minating in  the  base)  shall  be  =  DC  ;  then  daw  CO  and 
CH  parallel  to  FB  and  FA,  so  shall  GCH  i)e  the  triangle 
required. 

DEMONSTRATION. 
It  is  evident  that  the  angle  HCG  is  =  AFB  =  the 
given  one.  Moreover,  if  EM  a^d  EN  be  taken  as 
perpendiculars  to  AF  and  BF,  the/  will  be  equal  to  each 
other,  and  also  equal  to  the  given  one  AC,  because' 
all  the  angles  EFN,  EFM,  and  ADC  are  equal,  by 
construction,  and  EF  is  likewise  =  CD  ;  whence,  as 
the  angles  AHC,  AGC  are  respectively  equal  to  EAM, 


39B 


TJie  Construction  x>f 


EBN,  it  is  evident  that  HG  =  EA,  and  GC  =  EB,  and 
consequently  that  HC  +  GC  (  =  EA  +  EB)  =  AB* 

Method  of  Calculation. 

By  the  problem  above  referred  to,  AB  :  CD  (EF)  :  : 
co-sine  ADC  (AFE)  :  tangent  of  an  angle  ;  which  let 
be  denoted  by  Q. 

Now,  CD  :  CA  :  :  radius  :  sine  ADC  ;  which  pro- 
portion being  compounded  with  the  former,  we  have 
AB  :  CA  :  :  co-sine  ADC  x  radius  :  tangent  Q  x  sine 
A  -TN^       co-sine  ADC  \  radius    .  *  t^^, 

-^^^  =  -■  ^b^ADC (--tangent  ADC)  :  tau- 

gent  Q.  Then,  by  the  same  problem,  it  will  be  as  tangent 
^Q  :  radius  :  :  sine  ADC  :  co-sine  of  the  difference  of  the 
angles  (G  and  H)  at  the  base.  The  above  proportions, 
given  in  words  at  length,  exhibit  the  following  theorem: 

As  the  sum  of  the  sides  is  to  the  perpendicular^  so  is  the 
co-tangent  of  half  the  vertical  angle  to  the  tangent  of  an 
angle ;  and^  as  the  tangent  of  half  this  angle  is  to  the  ra- 
dius^ so  IS  the  sine  of  half  the  vertical  angle  to  the  cosine  of 
half  the  difference  of  the  angles  at  the  base. 


PROBLEM  LXXXI. 

To  constitute  a  trapezium  of  a  given  ?nagnitude  under 
four  given  lines* 

CONSTRUCTION. 
p/\^  Make   a  right 

P  /      ^^  angle  b  with  two 

aL       .^!"nn^  Qf  ^\^Q  given  lines 

Ab^  be  ;  and  with 
the  other  two 
complete  the  tra- 
pezium AbcT)  : 
upon  AD  let  fall 
the  perpendicular 
cE,  in  which  pro- 
duced (if  neces- 
sary) take  EF,  so 
that  the  rectangle 


Geometrical  Problems,.  399 

under  it,  and  AD,  may  be  double  the  given  area :  more- 
over, take  a  fourth  proportional  to  AD,  A^,  and  hc^  with 
which,  from  the  center  F,  let  an  arch  be  described,  meet- 
ing another  arch,  dt^scribed  from  D  with  the  radius  Dc  in 
C  ;  join  D,  C  ;  and  from  A  and  C  draw  the  other  two 
given  lines  AB,  CB,  so  as  to  meet,  and  they  will  thereby 
form  the  trapezium  A  BCD,  as  required. 

DEMONSTRATION. 

Draw  Ac,  AC,  and  FC  ;  upon  AD  and  AB  let  fall  the 
perpendiculars  CP,  CQ ;  and  make  FG  perpendicular  to 
PCG. 

Because  AD^  +  DC^  +  2AD  x  DP  (=  AC%  Euc. 
12.  2.)  =  AB2  +  BC2  +  2AB  X  BQ,  and  AD^ 
-f.  Dc^  +  2AD  X  DE  (=  Ac2)  =  A^^  ^  ^^.2  (^^^^ 
47.  1.),  it  follows,  by  taking  these  last  equal  quan- 
tities from  the  former,  that  2AD  x  DP  —  2AD  x 
DE  (2AD  -f-  EP)  =  2AB  x  BQ,  and  consequently 
that  BQ  ;  EP  (FG)  :  :  AD  :  AB  :  :  BC  :  FC  Qnj 
construction)  ;  whence  the  triangles  BCQ,  FCG  are 
similar,  and  so  CQ  :  CG  :  :  BC  :  FC  :  :  AD  :  AB  {bij 
construction)^  and  therefore  CQ  X  AB  =  CG  X  AD  ; 
hence,  by  adding  CP  x  AD  to  each,  we  have  CP  X  AD 
-j.  CQ  X  AB  (=  twice  the  area  ABCD)  =  CP  X  AD 
+  CG  X  AD  =  EF  X  AD  =  twice  the  given  area  (by 
construction).     ^  E.  Z). 

Method  of  Calculation. 

From  DE  (  =  A^^  +  ^c^        AD^  -  Dc-n  ^^^  ^^ 
\  2AD  / 

=  -T-TT-  )  the  value  of  DF,  and  likewise  that  of  the 
AD  / 

angle  ADF,  will  be  found  ;  then,  all  the  sides  of  the  tri- 
angle DCF  being  known,  the  angle  FDC  will  likewise  be 
known  ;  which,  added  to  ADF,  gives  (ADC)  one  of  the 
angles  of  the  trapezium. 

It  may  so  happen  tliat  a  trapeziui?i,  liaving  one  right 
angle,  cannot  be  constituted  under  the  four  given  lines  ; 
in  which  case  it  will  be  necessary  (instead  of  form- 
ing the  trapezium  A^cD)  to  lay  down  AD  first,  and 
in    it    {produced    if    needful)    to    take    DE    equal    to 


400  The  Construction^  &f  c. 

ABT  4- BCT — Ti3T— iDCV    ^      .  ,         , 

L_ 2AB ""      ^^  ^^^  ^^^      ^^ 

altitude  of  a  rectangle,  formed  on  the  base  2  AD,  whereof 
the  contained  area  is  equal  to  the  difference  of  AB^^  -f 
BUy  and  ADT  +"51:7  (which  line  DE  is  to  be  set  off 
on  the  other  side  of  D,  when  the  latter  of  these  two  quan- 
tities is  the  greater)  :  this  being  done,  the  rest  of  the  so- 
lution will  remain  the  same,  as  is  manifest  from  the  first 
and  second  steps  of  the  demonstration  ;  the  process  from 
thence  to  the  end  being  nowise  different. 

It  may  be  further  observed,  that  the  problem  itself  be- 
comes impossible,  when  the  two  circles,  described  from 
the  centers  D  and  F,  neither  cut  nor  touch  ;  the  greatest 
limit  of  the  area,  and  consequently  of  EF,  being  when  they 
touch  each  other ;  in  which  case  the  sum  of  the  radii 
DC,  FC  becoming  =  DF,  the  point  C  will  fall  in  the 
line  DF,  and  the  angle  DCF  will  become  equal  to  two 
right  angles  ;  but  the  sum  of  the  opposite  external  angles 
CDP  and  CBQ  is  always  equal  to  DCF  ;  because  CDP 
(supposing  Ctz  parallel  to  AP)  is  =  DC^z,  and  CBQ 
(  =  CFG)  =  FC/z :  hence  it  is  evident  that  the  limit,  or 
the  greatest  area  will  be  when  the  sum  of  the  opposite 
angles  is  equal  to  two  right  angles,  or  when  the  trapezium 
may  be  inscribed  in  a  circle. 


FINIS, 


Hi 


U  DAY  USE 

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